Question 11 Mark
Integrate the function: $\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$
AnswerLet $I=\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$
Put sin x = t ⇒ cos x dx = dt
$\Rightarrow \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{1}{\sqrt{4-t^{2}}} d t$
$=\int \frac{1}{\sqrt{\left(2^{2}-t^{2}\right)}} d t$
= $\sin ^{-1}\left(\frac{t}{2}\right)+C$
$\Rightarrow I=\sin ^{-1}\left(\frac{\sin x}{2}\right)+C$
View full question & answer→Question 21 Mark
Integrate the function $\int {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}} dx$
Answer$I=\int {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}} dx$ $ = \int {\frac{{{e^{\log {x^5}}} - {e^{\log {x^4}}}}}{{{e^{\log {x^3}}} - {e^{\log {x^2}}}}}} dx$
$=\int {\left( {\frac{{{x^5} - {x^4}}}{{{x^3} - {x^2}}}} \right)} dx$ $\left[ {\because {e^{\log \theta }} = \theta } \right]$
$ = \int {\frac{{{x^4}(x - 1)}}{{{x^2}(x - 1)}}} dx$
$ = \int {{x^2}dx} $
$ = \frac{{{x^3}}}{3} + c$
View full question & answer→Question 31 Mark
Integrate the function $\frac{\sin x}{\sin (x-a)}$
AnswerGiven Integrand is: $\frac{\sin x}{\sin (x-a)}$
Let $\mathrm{I}=\int\frac{\sin \mathrm{x}}{\sin (\mathrm{x}-\mathrm{a})}$
Let x - a = t $\Rightarrow$ x = t + a $\Rightarrow$ dx = dt
$\Rightarrow \int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (t+a)}{\sin (t)} d t$
As, {sin (A + B) = sin A cos B + cos A sin B}
$\Rightarrow \int \frac{\sin x}{\sin (x-a)} d x$ = $\int \frac{\sin t \cos a+\cos t \sin a}{\sin (t)} d t$
$=\int \frac{\sin t \cos a}{\sin t}+\frac{\cos t \sin a}{\sin t} d t$
= $\int(\cos a+\cot t \sin a) d t$
= $\int(\cos a) d t+\int(\cot t \sin a) d t$
= $\cos a \int 1 . \mathrm{dt}+\sin \mathrm{a} . \int(\cot t) \mathrm{d} \mathrm{t}$
= $(\cos a) \cdot(x-a)+\sin a \cdot \log |\sin (x-a)|+c$
= $\sin a \cdot \log |\sin (x-a)|+x \cdot \cos a-a \cdot \cos a+c$
= $\sin a \cdot \log |\sin (x-a)|+x \cdot \cos a+c_{1}$
View full question & answer→Question 41 Mark
Integrate the function $\frac{5 x}{(x+1)\left(x^{2}+9\right)}$
AnswerGiven: $\frac{5 x}{(x+1)\left(x^{2}+9\right)}$
Let $I=\frac{5 x}{(x+1)\left(x^{2}+9\right)}$
Using partial fraction:
Let $\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}$ ...(i)
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}$ = $\frac{A\left(x^{2}+9\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+9\right)}$
$\Rightarrow 5x = A(x^2+ 9) + (Bx + C)(x + 1)$
$\Rightarrow 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C$
$\Rightarrow 5x = 9A + C + (B + C)x + (A + B)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
9A + C = 0 $\Rightarrow$ C = -9A
B+C = 5 $\Rightarrow$ B = 5 - C $\Rightarrow$ B = 5 - (-9A) $\Rightarrow$ B = 5 + 9A
A + B = 0 ⇒ A = -B $\Rightarrow$ A = -(5 + 9A) $\Rightarrow$ 10A = -5 $\Rightarrow$ A = $\frac{-1}{2}$ and C = $\frac{9}{2}$ and B = $\frac{1}{2}$
Put these values in equation (i)
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}$
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{-\frac{1}{2}}{(x+1)}+\frac{\left(\frac{1}{2}\right) x+\frac{9}{2}}{\left(x^{2}+9\right)}$
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=-\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \cdot\left(\frac{x+9}{\left(x^{2}+9\right)}\right)$
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x$ = $-\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+\frac{1}{2} \cdot \int \frac{x}{\left(x^{2}+9\right)} d x+\frac{9}{2} \int \frac{1}{\left(x^{2}+9\right)} d x$
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x$ = $-\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+I_{1}+\frac{9}{2} \int \frac{1}{\left(x^{2}+\left(3^{2}\right))\right.} d x$
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x$ = $-\frac{1}{2} \cdot \log |\mathrm{x}+1|+\mathrm{I}_{1}+\frac{9}{2} \cdot\left(\frac{1}{3} \tan ^{-1} \frac{\mathrm{x}}{3}\right)$ ...(ii)
First solve for $I_1$:
$I_{1}=\frac{1}{2} \cdot \int \frac{x}{\left(x^{2}+9\right)} d x$
Put $x^2 = t \Rightarrow 2xdx = dt$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2} \cdot \int \frac{1}{(\mathrm{t}+9)} \cdot \frac{\mathrm{dt}}{2}$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2} \cdot \int \frac{1}{(\mathrm{t}+9)} \cdot \frac{\mathrm{dt}}{2}$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{4} \log \left|\mathrm{x}^{2}+9\right|$
Put the value in equ. (ii)
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x$ = $-\frac{1}{2} \cdot \log |\mathrm{x}+1|+\frac{1}{4} \log \left|\mathrm{x}^{2}+9\right|+\frac{3}{2} \cdot\left(\tan ^{-1} \frac{\mathrm{x}}{3}\right)+\mathrm{C}$
View full question & answer→Question 51 Mark
Integrate the function $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ [Hint: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$ Put $x = t^6$]
AnswerGiven: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ or we can write it as $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$
Let $x = t^6 \Rightarrow dx = 6t^5dt$
$\Rightarrow \int \frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} \cdot d x=\int \frac{6 t^{5}}{t^{2}(1+t)} \cdot d t$
= $\text { 6. } \int \frac{t^{3}}{(1+t)} \cdot d t$
After division we get,
$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ = $\int\left[\left(t^{2}-t+1\right)-\frac{1}{(1+t)}\right] \cdot d t$
= $6 .\left\{\int \mathrm{t}^{2} . \mathrm{dt}-\int \mathrm{t} . \mathrm{dt}+\int 1 . \mathrm{dt}-\int\left[\frac{1}{(1+t)}\right] \cdot \mathrm{dt}\right\}$
= $6\left[\left(\frac{t^{3}}{3}\right)-\left(\frac{t^{2}}{2}\right)+t-\log (1+t)\right]$
= $6\left[\left(\frac{\left(x^{\frac{1}{6}}\right)^{3}}{3}\right)-\left(\frac{\left(x^{\frac{1}{6}}\right)^{2}}{2}\right)+\left(x^{\frac{1}{6}}\right)-\log \left(1+\left(x^{\frac{1}{6}}\right)\right)\right]+c$
= $\left[\left(2 x^{\frac{1}{2}}\right)-\left(3 x^{\frac{1}{3}}\right)+6 \cdot x^{\frac{1}{6}}-6 \cdot \log \left(1+x^{\frac{1}{6}}\right)\right]+C$
= $2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+c$
View full question & answer→Question 61 Mark
If f(a + b - x) = f (x), then $\int_{a}^{b} x f(x) d x$ is equal to
AnswerGiven Integral is: $\int_{a}^{b} x f(x) d x$
Let $\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
as, f(x) = f(a + b - x)
$\Rightarrow \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}(\mathrm{a}+\mathrm{b}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}-\int_{\mathrm{a}}^{\mathrm{b}}(\mathrm{x}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{2}^{\mathrm{b}}(\mathrm{a}+\mathrm{b}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}-\mathrm{I}$
$\Rightarrow 2 \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}(\mathrm{a}+\mathrm{b}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\frac{(\mathrm{a}+\mathrm{b})}{2} \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
View full question & answer→Question 71 Mark
Integrate the function $\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$
AnswerLet $I=\frac{1}{x^{2} \cdot\left(x^{4}+1\right)^{\frac{3}{4}}}$
Taking $x^{4}$ common from the denominator, we get
$I =\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$=\int \frac{d x}{x^{2}\left(x^{3}\right)\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$=\int \frac{d x}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$\text { Let } t=1+\frac{1}{x^{4}} \Rightarrow-\frac{d t}{4}=\frac{d x}{x^{5}}$
$\Rightarrow \int \frac{1}{x^{2} \cdot\left(x^{4}+1\right)^{\frac{3}{4}}} \cdot d x$ $=\frac{-1}{4} \int \frac{d t}{t^{\frac{3}{4}}}$
= $\frac{-1}{4}\left(\frac{t^{\frac{1}{4}}}{\frac{1}{4}}\right)+C$
= $-t^{\frac{1}{4}}+c$
=$-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$
View full question & answer→Question 81 Mark
$\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is equal to
AnswerGiven Integral is: $\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$
Let $I=\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$
= $\int \frac{\cos ^{2} x-\sin ^{2} x}{(\sin x+\cos x)^{2}} d x$
= $\int \frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^{2}} d x$
= $\int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x$
Put sin x + cos x = t $\Rightarrow$ (cos x - sin x)dx = dt
$\Rightarrow \int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x=\int \frac{d t}{t}$
= log |t| + C
= log |sin x + cos x| + C
View full question & answer→Question 91 Mark
$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to
AnswerGiven Integral is: $\int \frac{d x}{e^{x}+e^{-x}}$
Let $I=\int \frac{d x}{e^{x}+e^{-x}}$
= $\int \frac{d x}{e^{-x}\left(e^{2 x}+1\right)}$
= $\int \frac{e^{x} d x}{\left(e^{2 x}+1\right)}$
Put $e^x = t \Rightarrow e^x$ dx = dt
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}}{\left(\mathrm{e}^{2 \mathrm{x}}+1\right)}=\int \frac{\mathrm{dt}}{\left(\mathrm{t}^{2}+1\right)}$
$= \tan^{-1} t + C$
$= \tan^{-1} (e^x) + C$
View full question & answer→Question 101 Mark
Prove $\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1$
AnswerGiven integral is: $\int_{0}^{1} \sin ^{-1} x d x$
To Prove: $\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1$
Let $I=\int_{0}^{1} \sin ^{-1} x \cdot 1 d x$
because, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} \cdot \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{d}}{\mathrm{dx}} u \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow I$ = $\sin ^{-1} \mathrm{x} \cdot \int_{0}^{1} 1 \cdot \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x} \cdot\left\{\int 1 \cdot \mathrm{dx}\right\} \cdot \mathrm{dx}$
$\Rightarrow I=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \cdot \mathrm{x} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-\mathrm{I}_1$ .....(i)
First solving $I_1$:
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \cdot \mathrm{x} \mathrm{d} \mathrm{x}$
Let $1 - x^2 = t \Rightarrow -2 x dx = dt$
When x = 0 then t = 1 and when x = 1 then t = 0
$\Rightarrow \mathrm{I}_{1}=\int_{1}^{0} \frac{1}{\sqrt{\mathrm{t}}} \cdot \frac{-\mathrm{dt}}{2}$
= $-\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]_{1}^{0}$
$\Rightarrow \mathrm{I}_{1}=\sqrt{1}$
$\Rightarrow \mathrm{I}_{1}=1$
Put in equation (i)
$\Rightarrow \mathrm{I}=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-1$
$\Rightarrow \mathrm{I}=\sin ^{-1}(1)-0-1$
$\Rightarrow \mathrm{I}=\frac{\pi}{2}-1$
Hence Proved.
View full question & answer→Question 111 Mark
Prove $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$
AnswerGiven integral is: $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x$
To Prove: $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$
Let $I=\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x$ ...(i)
= $\int_{0}^{\frac{\pi}{4}} 2 \cdot \tan x \cdot \tan ^{2} x d x$
= $\text { 2. } \int_{0}^{\frac{\pi}{4}} \tan x \cdot\left(\sec ^{2} x-1\right) d x$
$\Rightarrow \mathrm{I}=2\left\{-\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \mathrm{dx}+\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}\right\}$
$\Rightarrow \mathrm{I}=-[2 \log \cos \mathrm{x}]_{0}^{\pi / 4}+2 . \mathrm{I}_{1}$ ...(ii)
Solving $I_1$:
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$
Let, tan $x = t \Rightarrow \sec^2 x dx = dt$
When x = 0 then t = 0 and when x = $\frac{\pi}{4}$ then t = 1
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \mathrm{t} \mathrm{dt}$
= $\left[\frac{t^{2}}{2}\right]_{0}^{1}$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2}$
Using this in equation (ii)
$\Rightarrow \mathrm{I}=[2 \log \cos \mathrm{x}]_{0}^{\pi / 4}+2 \cdot \frac{1}{2}$
$\Rightarrow \mathrm{I}=2\left\{\log \cos \frac{\pi}{4}-\log \cos 0\right\}+1$
$\Rightarrow \mathrm{I}=2\left\{\log \frac{1}{\sqrt{2}}-\log 1\right\}+1$
$\Rightarrow \mathrm{I}=\left\{\log \left(\frac{1}{\sqrt{2}}\right)^{2}-\log (1)^{2}\right\}+1$
$\Rightarrow \mathrm{I}=1-\log 2+\log 1$
$\Rightarrow I=1-\log 2$
Hence Proved.
View full question & answer→Question 121 Mark
Prove $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
AnswerGiven integral is: $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$
To prove: $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
Let $I=\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$ ...(i)
= $\int_{0}^{\frac{\pi}{2}} \sin x \cdot \sin ^{2} x d x$
= $\int_{0}^{\frac{\pi}{2}} \sin x \cdot\left(1-\cos ^{2} x\right) d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \mathrm{dx}-\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x} d \mathrm{x}$
$\Rightarrow \mathrm{I}=[-\cos \mathrm{x}]_{0}^{\pi / 2}-\mathrm{I}_{1}$ ...(ii)
Now, we solve $I_1$:
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$
Let cos x = t $\Rightarrow$ -sin x dx = dt $\Rightarrow$ sinx dx = -dt
When x = 0 then t = 1 and when x = $\frac{\pi}{2}$ then t = 0
$\Rightarrow \mathrm{I}_{1}=\int_{1}^{0} \mathrm{t}^{2}(-\mathrm{dt})$
= $-\int_{1}^{0} t^{2}(d t)$
= $-\left[\frac{t^{3}}{3}\right]_{1}^{0}$
= $-\left\{-\frac{1}{3}\right\}$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{3}$
Using this value in equation (ii)
$\Rightarrow \mathrm{I}=[-\cos \mathrm{x}]_{0}^{\pi / 2}-\frac{1}{3}$
$\Rightarrow \mathrm{I}=-\left\{\cos \frac{\pi}{2}-\cos 0\right\}-\frac{1}{3}$
$\Rightarrow \mathrm{I}=1-\frac{1}{3}$
$\Rightarrow \mathrm{I}=\frac{2}{3}$
Hence Proved.
View full question & answer→Question 131 Mark
Prove $\int_{-1}^{1} x^{17} \cos ^{4} x d x=0$
AnswerLet $\mathrm{I}=\int_{-1}^{1} \mathrm{x}^{17} \cdot \cos ^{4} \mathrm{xdx}$
As we can see $f(x) =x^{17}.\cos^4x$ and $f(-x) = (-x)^{17}.\cos^4(-x) = -x^{17}.\cos^4x$
i.e. $f(x) = -f(-x)$
so, it is an odd function.
It is also known that if f(x) is an odd function then $\left\{\int_{-a}^{a} f(x) d x=0\right\}$
$\Rightarrow \mathrm{I}=\int_{-1}^{1} \mathrm{x}^{17} \cdot \cos ^{4} \mathrm{xdx}=0$
Hence proved.
View full question & answer→Question 141 Mark
Prove $\int_{0}^{1} x e^{x} d x=1$
AnswerTo Prove: $\int_{0}^{1} x e^{x} d x=1$
Let $\mathrm{I}=\int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
because, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} . \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{dx}=[\mathrm{x}]_{0}^{1} \cdot \int_{0}^{1} \mathrm{e}^{\mathrm{x}} \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{d}}{\mathrm{dx}}x \cdot\left\{\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}\right\} \cdot \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 \cdot e^{x} d x$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{1}-e^{0}\right]$
$\Rightarrow \int_{0}^{1} x e^{x} d x=e-0-e+1$
$\Rightarrow \int_{0}^{1} x e^{x} d x=1$
Hence Proved.
View full question & answer→Question 151 Mark
Prove that $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
AnswerGiven integral is: $\int_{1}^{3} \frac{d x}{x^{2} (x+1)}$
To Prove: $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
Let $I=\frac{d x}{\left(x^{2}\right)(x+1)}$
Using partial fraction:
Let $\frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$ .....(i)
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}$ = $\frac{A(x)(x+1)+B(x+1)+C\left(x^{2}\right)}{(x+1)\left(x^{2}\right)}$
$\Rightarrow 1=\mathrm{A}\left(\mathrm{x}^{2}+\mathrm{x}\right)+(\mathrm{Bx}+\mathrm{B})+\mathrm{Cx}^{2}$
$\Rightarrow 1 = Ax^2 + Ax + B + Bx + Cx^2$
$\Rightarrow 1 = B + (A + B)x + (A + C)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
B = 1
A + B = 0 $\Rightarrow$ A = -B $\Rightarrow$ A = -1
A + C = 0 $\Rightarrow$ C = -A $\Rightarrow$ C = 1
Put these values in equation (i)
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}$
$\Rightarrow \int \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\int-\frac{1}{x} d x+\int \frac{1}{\left(x^{2}\right)} d x+\int \frac{1}{(x+1)} d x$
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\log |x|-x^{-1}+\log |x+1|\right]_{1}^{3}$
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\frac{1}{x}+\log \left|\frac{x+1}{x}\right|\right]_{1}^{3}$
= $\left[-\frac{1}{3}+\log \left|\frac{3+1}{3}\right|-\left(-\frac{1}{1}+\log \left|\frac{1+1}{1}\right|\right)\right]$
= $\left[-\frac{1}{3}+\log \left|\frac{4}{3}\right|+\left(1-\log \left|\frac{2}{1}\right|\right)\right]$
= $\left[-\frac{1}{3}+1+\log \left|\frac{4}{3} \times \frac{1}{2}\right|\right]$
$\Rightarrow \mathrm{I}=\left[\frac{2}{3}+\log | \frac{2}{3}\right] |$
L.H.S = R.H.S
Hence proved.
View full question & answer→Question 161 Mark
Evaluate the definite integral $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$
AnswerGiven: $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$
$\Rightarrow \mathrm{I}=\int_{1}^{4}[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|] \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{1}^{4}[|\mathrm{x}-1|] \mathrm{dx}+\int_{1}^{4}[|\mathrm{x}-2|] \mathrm{d} \mathrm{x}+\int_{1}^{4}[|\mathrm{x}-3|] \mathrm{d} \mathrm{x}$
Let $I = I_1 + I_2 + I_3$
First solve for $I_1$:
$I_{1}=\int_{1}^{4}[|x-1|] d x$
As we can see that $(x-1) \geq 0$ when $1 \leq x \leq 4$
$\Rightarrow \mathrm{I}_{1}=\int_{1}^{4}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{1}=\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{1}^{4}$
$\Rightarrow \mathrm{I}_{1}=\left[\frac{(4)^{2}}{2}-4-\frac{(1)^{2}}{2}+1\right]$
$\Rightarrow \mathrm{I}_{1}=\left[8-4-\frac{1}{2}+1\right]$
$\Rightarrow I_{1}=\left[5-\frac{1}{2}\right]$
$\Rightarrow \mathrm{I}_{1}=\frac{9}{2}$
Now solve for $I_2$:
$I_{2}=\int_{1}^{4}[|x-2|] d x$
As we can see that $(x-2) \leq 0$ when $1 \leq x \leq 2$ and $(x-2) \geq 0$ when $2 \leq x \leq 4$
as, $\left\{\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\right\}$
$\Rightarrow \mathrm{I}_{2}=\int_{1}^{2}-(\mathrm{x}-2) \mathrm{d} \mathrm{x}+\int_{2}^{4}(\mathrm{x}-2) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{\mathrm{x}^{2}}{2}-2 \mathrm{x}\right]_{1}^{2}+\left[\frac{\mathrm{x}^{2}}{2}-2 \mathrm{x}\right]_{2}^{4}$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{(2)^{2}}{2}-2(2)-\frac{(1)^{2}}{2}+2(1)\right]$ + $\left[\frac{(4)^{2}}{2}-2(4)-\frac{(2)^{2}}{2}+2(2)\right]$
$\Rightarrow \mathrm{I}_{2}=-\left[2-4-\frac{1}{2}+2\right]+[8-8-2+4]$
$\Rightarrow \mathrm{I}_{2}=\left[\frac{1}{2}+2\right]$
$\Rightarrow \mathrm{I}_{2}=\frac{5}{2}$
Now solve for $I_3$:
$I_{3}=\int_{1}^{4}[|x-3|] d x$
As we can see that $(x-3) \leq 0$ when $1 \leq x \leq 3$ and $(x-3) \geq 0$ when $3 \leq x \leq 4$
as, $\left\{\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\right\}$
$\Rightarrow \mathrm{I}_{3}=\int_{1}^{3}-(\mathrm{x}-3) \mathrm{d} \mathrm{x}+\int_{3}^{4}(\mathrm{x}-3) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{1}^{3}+\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{3}^{4}$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{(3)^{2}}{2}-3(3)-\frac{(1)^{2}}{2}+3(1)\right]$ + $\left[\frac{(4)^{2}}{2}-3(4)-\frac{(3)^{2}}{2}+3(3)\right]$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{9}{2}-9-\frac{1}{2}+3\right]+\left[8-12-\frac{9}{2}+9\right]$
$\Rightarrow I_{3}=\left[2+\frac{1}{2}\right]$
$\Rightarrow I_{3}=\frac{5}{2}$
as $I = I_1 + I_2 + I_3$
$\Rightarrow I=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}$
$\Rightarrow \mathrm{I}=\frac{19}{2}$
View full question & answer→Question 171 Mark
Evaluate the definite integral $\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$
AnswerAccording to the question , $I =\int _ { 0 } ^ { \pi / 2 } 2 \sin x \cos x \tan ^ { - 1 } ( \sin x ) d x$
Put , sin x = t
$\Rightarrow$ cos x dx = dt
Lower Limit , when x= 0, then t = sin 0 =0
Upper Limit , when x =$\frac{\pi}{2}$ , then t = sin$\frac{\pi}{2}$ = 1.
$\therefore \quad I = 2 \int _ { 0 } ^ { 1 } t \times \tan ^ { - 1 } t d t$
Applying integration by parts,we get
$I = 2 \left[ \frac { t ^ { 2 } } { 2 } \times \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } - 2 \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + t ^ { 2 } } \times \frac { t ^ { 2 } } { 2 } d t$
$= 2 \left[ \frac { t ^ { 2 } } { 2 } \times \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \frac { t ^ { 2 } } { 1 + t ^ { 2 } } d t$
$= 2 \times \frac { 1 } { 2 } \times \tan ^ { - 1 } ( 1 ) - \int _ { 0 } ^ { 1 } \frac { 1 + t ^ { 2 } - 1 } { 1 + t ^ { 2 } } d t$
$= 1 \times \frac { \pi } { 4 } - \int _ { 0 } ^ { 1 } \left( \frac { 1 + t ^ { 2 } } { 1 + t ^ { 2 } } - \frac { 1 } { 1 + t ^ { 2 } } \right) d t$
$= \frac { \pi } { 4 } - \int _ { 0 } ^ { 1 } \left( 1 - \frac { 1 } { 1 + t ^ { 2 } } \right) d t$
$= \frac { \pi } { 4 } - \left[ t - \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 }$
$= \frac { \pi } { 4 } - 1 + \tan ^ { - 1 } 1$
$= \frac { \pi } { 4 } - 1 + \frac { \pi } { 4 } = \frac { 2 \pi } { 4 } - 1$
$ = \frac { \pi } { 2 } - 1 $
$\therefore \quad I = \frac { \pi } { 2 } - 1 $
View full question & answer→Question 181 Mark
Integrate the function $\frac{1}{x \sqrt{a x-x^{2}}}$ [Hint: Put x = $\frac{a}{t}$]
AnswerGiven: $\frac{1}{x \sqrt{a x-x^{2}}}$
Let $I=\frac{1}{x \sqrt{a x-x^{2}}}$
Put $x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^{2}} d t$
$\Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{\frac{a \cdot a}{t}-\left(\frac{a}{t}\right)^{2}}} \cdot-\frac{a}{t^{2}} d t$
$=\int \frac{-1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\left(\frac{1}{t}\right)^{2}}} d t$
= $-\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^{2}}{t}-\left(\frac{t}{t}\right)^{2}}} d t$
= $-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t$
= $-\frac{1}{a} \int(t-1)^{-\frac{1}{2}} d t$
= $-\frac{1}{a}\left[\frac{\sqrt{(t-1)}}{\frac{1}{2}}\right]+C$
= $-\frac{2}{a}[\sqrt{\left(\frac{a}{x}-1\right)}]+C$ ; because $t=\frac{a}{x}$
$\Rightarrow \mathrm{I}=-\frac{2}{\mathrm{a}}[\sqrt{\left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{x}}\right)}]+\mathrm{C}$
View full question & answer→Question 191 Mark
Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$
AnswerAccording to the question, $I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \sin 2 x } d x$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 ( 1 + \sin 2 x - 1 ) } d x$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 [ 1 - ( 1 - \sin 2 x ) ] } d x$
$\Rightarrow I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \left[ \begin{array} { r } { 1 - \left( \cos ^ { 2 } x + \sin ^ { 2 } x \right. } { - 2 \sin x \cos x ) } \end{array} \right] } d x$$[\because 1 = \cos ^ { 2 } x + \sin ^ { 2 } x ]$ and $[\because \sin 2 x = 2 \sin x \cos x ]$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \left[ 1 - ( \cos x - \sin x ) ^ { 2 } \right] } d x$
put , $cos x - sin x = t$
$\Rightarrow (- sin x - cos x) dx = dt$
$\Rightarrow (sin x + cos x) dx = - dt$
Lower limit , when x = 0, then t = cos 0 - sin 0 = 1
Upper limit , when x = $\frac { \pi } { 4 } , $ then $t = \cos \frac { \pi } { 4 } - \sin \frac { \pi } { 4 } = \frac { 1 } { \sqrt { 2 } } - \frac { 1 } { \sqrt { 2 } } = 0.$
$\therefore \quad I = \int _ { 1 } ^ { 0 } \frac { - d t } { 9 + 16 \left( 1 - t ^ { 2 } \right) }$
$\Rightarrow \quad I = \int _ { 0 } ^ { 1 } \frac { d t } { 9 + 16 \left( 1 - t ^ { 2 } \right) }$
$= \int _ { 0 } ^ { 1 } \frac { d t } { 25 - 16 t ^ { 2 } }$
$= \frac { 1 } { 16 } \int _ { 0 } ^ { 1 } \frac { d t } { \left( \frac { 5 } { 4 } \right) ^ { 2 } - t ^ { 2 } }$
$= \frac { 1 } { 2 \times \frac { 5 } { 4 } \times 16 } \left[ \log \left| \frac { 5 + 4 t } { 5 - 4 t } \right| \right] _ { 0 } ^ { 1 }$$\left[ \because \int \frac { 1 } { a ^ { 2 } - x ^ { 2 } } d x = \frac { 1 } { 2 a } \log \left| \frac { a + x } { a - x } \right| + C \right]$
$= \frac { 1 } { 40 } \left[ \log \left| \frac { 5 + 4 } { 5 - 4 } \right| - \log \left| \frac { 5 } { 5 } \right| \right]$
$= \frac { 1 } { 40 } \left[ \log \left( \frac { 9 } { 1 } \right) - \log \left( \frac { 5 } { 5 } \right) \right]$
$= \frac { 1 } { 40 } ( \log 9 - \log 1 ) $
$= \frac { 1 } { 40 } ( \log 9 )$ [$ \because$ log 1 = 0]
$\Rightarrow I = \frac { 1 } { 40 } \log ( 3 ) ^ { 2 }$
$= \frac { 2 } { 40 } \log 3$ [$\because$log $a^n$ = nlog a]
$\therefore \quad I = \frac { 1 } { 20 } \log 3$
View full question & answer→Question 201 Mark
Evaluate the definite integral $\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
AnswerIntegral to be evaluated: $\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
Let $I=\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
= $\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} d x$
= $\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x$
= $\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1} d x$
= $\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x$
= $\int_{0}^{1}\left((1+x)^{\frac{1}{2}}\right) d x+\int_{0}^{1}(x)^{\frac{1}{2}} d x$
$\Rightarrow \mathrm{I}=\left[\frac{(1+\mathrm{x})^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{(\mathrm{x})^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}$
= $\frac{2}{3} \cdot\left[(1+1)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]$
= $\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]$
= $\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-1\right]+\frac{2}{3} \cdot[1]$
= $\frac{2}{3} \cdot[2 \sqrt{2}]$
$\Rightarrow$ $I=\frac{4 \sqrt{2}}{3}$
View full question & answer→Question 211 Mark
Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
Answer$I = \int_{\pi /6}^{\pi /3} {\frac{{\sin x + \cos x}}{{\sqrt {\sin 2x} }}dx} $put sinx - cox = t
$(cosx + sinx)dx = dt$
$(sinx - cosx)^2 = t^2$
$1 - \sin^2x = t^2$
$\sin 2x = 1 - t^2$
when $x = \frac{\pi }{6}$
$t = \frac{1}{2} - \frac{{\sqrt 3 }}{2}$
When $x = \frac{\pi }{3}$
$t = \frac{{\sqrt 3 }}{2} - \frac{1}{2}$
I=$\int_{\frac{{1 - \sqrt 3 }}{2}}^{\frac{{\sqrt 3 - 1}}{2}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} $
$= \left[ {{{\sin }^{ - 1}}t} \right]_{\frac{{1 - \sqrt 3 }}{2}}^{\frac{{\sqrt 3 - 1}}{2}}$
$= {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{{1 - \sqrt 3 }}{2}} \right)$
$= {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right)$
$= 2{\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right)$
View full question & answer→Question 221 Mark
Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}$
AnswerGiven $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$ ......(i)
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} \mathrm{x}}{\cos ^{2} \mathrm{x}+4\left(1-\cos ^{2} \mathrm{x}\right)} \mathrm{d} \mathrm{x}$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4(1)-\left(4 \cos ^{2} x\right)} d x$
= $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{4-3 \cos ^{2} x} d x$
= $\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{3} \cdot 3 \cos ^{2} x}{4-3 \cos ^{2} x} d x$
= $-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{-3 \cos ^{2} x+4-4}{4-3 \cos ^{2} x} d x$
= $-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4}{4-3 \cos ^{2} x} d x$
= $-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}}(1) d x+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4}{4-3\left(\frac{1}{\sec ^{2} x}\right)} d x$
= $-\frac{1}{3} \cdot[\mathrm{x}]_{0}^{\frac{\pi}{2}}+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4 \sec ^{2} \mathrm{x}}{4 \sec ^{2} \mathrm{x}-3} \mathrm{dx}$
= $-\frac{1}{3} \cdot\left[\frac{\pi}{2}\right]+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4\left(1+\tan ^{2} x\right)-3} d x$
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} \mathrm{x}}{1+4 \tan ^{2} \mathrm{x}} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\mathrm{I}_{1}$ .....(ii)
First solve for $I_1$:
$I_{1}=\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$
Let 2 tan x = t $\Rightarrow$ 2 $\sec^2$ x dx dt
When x = 0 then t = 0 and when x = $\frac{\pi}{2}$ then t = $\infty$
$\Rightarrow \frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\frac{2}{3} \cdot \int_{0}^{\infty} \frac{1}{1+t^{2}} d t$
$\Rightarrow \mathrm{I}_{1}=\frac{2}{3}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{\infty}$
= $\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]$
$\Rightarrow \mathrm{I}_{1}=\frac{2}{3} \cdot \frac{\pi}{2}$
$\Rightarrow \mathrm{I}_{1}=\frac{\pi}{3}$
Put this value in equ.(ii)
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{\pi}{3}$
$\Rightarrow \mathrm{I}=\frac{\pi}{6}$
View full question & answer→Question 231 Mark
Evaluate the definite integral$\int_0^{\frac{\pi }{4}} {\frac{{\sin x \cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}} dx$
Answer$I = \int_0^{\pi /4} {\frac{{\sin x.\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} $Dividing Nr. and Dr. by $\cos^4x$
$ = \int_0^{\pi /4} {\frac{{\frac{{\sin x.\cos x}}{cos^4x}}}{{\frac{{{{\cos }^4}x}}{{{{\cos }^4}x}} + \frac{{{{\sin }^4}x}}{{{{\cos }^4}x}}}}dx} $
$ = \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{\tan }^4}x}}dx} $
$ = \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{({{\tan }^2}x)}^2}}}dx} $
Put ${\tan ^2}x = t$
$2\tan x.{\sec ^2}xdx = dt$
When x = 0, t= 0 and when $x=\frac{π}{4},t=1$
$\therefore{I} = \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 + {t^2}}}} $
$ = \frac{1}{2}\left[ {{{\tan }^{ - 1}}t} \right]_0^1$
$ = \frac{1}{2}.\frac{\pi }{4} = \frac{\pi }{8}$
View full question & answer→Question 241 Mark
Evaluate the definite integral $\int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$
AnswerLet $I=\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\right.$
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x\right.$
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2 \sin ^{2}\left(\frac{x}{2}\right)}-\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x\right.$
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x\right.$
Now let $f(x)=-\cot \frac{x}{2}$
$\Rightarrow$ f'(x) = $-\left(-\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)\right)=\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)$
$\Rightarrow \int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x\right.$ = $\int_{-\frac{\pi}{2}}^{\pi}\left(f(x)+f^{\prime}(x)\right) e^{x} d x$
= $\left[e^{x} f(x)\right]_{-\frac{\pi}{2}}^{\pi}$
= $\left[e^{x}\left(-\cot \frac{x}{2}\right)\right]_{-\frac{\pi}{2}}^{\pi}$
= $-\left[e^{\pi}\left(\cot \frac{\pi}{2}\right)-e^{\frac{\pi}{2}}\left(\cot \frac{\pi}{4}\right)\right]$
= $-\left[e^{\pi}(0)-e^{\frac{\pi}{2}}(1)\right]$
= $-\left[0-e^{\frac{\pi}{2}}\right]$
$\Rightarrow I=e^{\frac{\pi}{2}}$
View full question & answer→Question 251 Mark
Integrate the function $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$
AnswerGiven integrand is: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$
Here we can rewrite integrand as; $ \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$
= $\frac{\sqrt{x^{2}+1}}{x^{4}}\left[\log \left(x^{2}+1\right)-\log x^{2}\right]$
= $\frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}}\left[\log \left(\frac{x^{2}+1}{x^{2}}\right)\right]$
= $\frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}}\left[\log \left(1+\frac{1}{x^{2}}\right)\right]$
Let, $1+\frac{1}{x^{2}}=t \Rightarrow-\frac{2}{x^{3}} d x=d t$
Now, Let $I = \int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} d x$ = $\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}}\left[\log \left(1+\frac{1}{x^{2}}\right)\right] d x$
= $\int-\frac{1}{2} \cdot \sqrt{t}[\log (t)] d t$
= $\int-\frac{1}{2} \cdot \sqrt{t}[\log (t)] d t$ = $-\frac{1}{2}\left[\log \mathrm{t} . \int \sqrt{\mathrm{t}} \mathrm{d} \mathrm{t}-\int \frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t} .\left\{\int \sqrt{\mathrm{t}} \mathrm{d} \mathrm{t}\right\} \mathrm{dt}\right]$
= $-\frac{1}{2}\left[\log \mathrm{t} \cdot \frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{\mathrm{t}} \cdot\left\{\frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}\right\} \mathrm{dt}\right]$
= $-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\int\left\{\frac{t^{\frac{1}{2}}-1}{\frac{3}{2}}\right\} d t\right]$
= $-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \int t^{\frac{1}{2}} d t\right]$
= $-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]$
= $\left[-\frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} \log t+\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}}\right]$
= $-\frac{1}{3} t^{\frac{3}{2}}\left[\log t-\frac{2}{3}\right]$
$\Rightarrow 1=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left[\log \left(1+\frac{1}{x^{2}}\right)-\frac{2}{3}\right]+C$
View full question & answer→Question 261 Mark
Integrate the function $\text { tan }^{-1} \sqrt{\frac{1-x}{1+x}}$
AnswerLet $I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$
Let x = cos $\theta$ $\Rightarrow$ dx = -sin $\theta$ d$\theta$
$\Rightarrow \theta=\cos ^{-1} x$
$\Rightarrow \mathrm{I}=\int \tan ^{-1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{d} \mathrm{x}$ = $\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta) d \theta$
= $-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}}(\sin \theta) d \theta$
= $-\int \tan ^{-1} \sqrt{\tan ^{2}\left(\frac{\theta}{2}\right)}(\sin \theta) d \theta$
= $-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot(\sin \theta) d \theta$
= $-\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta$
As we know, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} . \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{d}}{\mathrm{dx}}(u) \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow -\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta$ = $-\frac{1}{2}\left[\theta . \int \sin \theta \mathrm{d} \theta-\int \frac{\mathrm{d}}{d \theta}{(\theta)} \cdot\left\{\int \sin \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]$
= $-\frac{1}{2}\left[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right]$
= $-\frac{1}{2}[-\theta \cdot \cos \theta+\sin \theta]$
= $\frac{1}{2} \theta \cdot \cos \theta-\frac{1}{2} \sqrt{\left(1-\cos ^{2} \theta\right.})$
= $\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{\left(1-x^{2}\right.}+C$
= $\frac{1}{2}\left(x . \cos ^{-1} x-\sqrt{\left(1-x^{2}\right.}\right)+C$
View full question & answer→Question 271 Mark
Integrate the function $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
AnswerGiven: $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
Let $I=\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
Using partial fraction:
Let $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}$ .....(i)
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$ = $\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^{2}}{(x+1)^{2}(x+2)}$
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$ = $\frac{A\left(x^{2}+3 x+2\right)+B(x+2)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}(x+2)}$
$\Rightarrow x^2 + x + 1 = Ax^2 + 3Ax + 2A + Bx +2B + Cx^2 + 2Cx + C$
$\Rightarrow x^2 + x + 1 = (2A + 2B + C) + (3A + B + 2C)x + (A + C)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
A + C = 1
3A + B + 2C = 1
2A + 2B + C = 1
After solving we get:
A = -2, B = 1 and C = 3
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}$
$\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ = $\int\left(\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right) d x$
= $-2 . \int\left(\frac{1}{x+1}\right) d x+\int\left(\frac{1}{(x+1)^{2}}\right) d x+3 . \int\left(\frac{1}{(x+2)}\right) d x$
= $-2 \cdot \int\left(\frac{1}{x+1}\right) d x+\int\left((x+1)^{-2}\right) d x+3 \cdot \int\left(\frac{1}{(x+2)}\right) d x$
= $-2 \log |x+1|+\left(\frac{(x+1)^{-1}}{(-1)}\right)+3 \log |x+1|+C$
= $-2 \log |x+1|-\frac{1}{(x+1)}+3 \log |x+1|+C$
View full question & answer→Question 281 Mark
Integrate the function $\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}$
AnswerLet $I=\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}$
= $\left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right) e^{x}$
= $2 \cdot\left(\frac{1+\sin x \cos x}{2 \cos ^{2} x}\right) e^{x}$
= $\left(\frac{1}{\cos ^{2} x}+\frac{\sin x \cos x}{\cos ^{2} x}\right) e^{x}$
= $\left(\sec ^{2} x+\tan x\right) e^{x}$
$\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x=\int\left(\sec ^{2} x+\tan x\right) e^{x} d x$
Now let tan x = f(x)
$\Rightarrow f'(x) = \sec^2x dx$
$\Rightarrow \int\left(\sec ^{2} x+\tan x\right) e^{x} d x = \int\left(f(x)+f^{\prime}(x)\right) e^{x} d x = e^{x} f(x)+C$
$\Rightarrow I=e^{x} \tan x+C$
View full question & answer→Question 291 Mark
Integrate the function: $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$
AnswerLet I = $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$. Then,
$I=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} \times d x$
$=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{x+a-x-b} \times d x$
$=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} \times d x$
$=\frac{1}{a-b}\left[\frac{2}{3}(x+a)^{\frac{3}{2}}-\frac{2}{3}(x+b)^{\frac{3}{2}}\right]+c$
$=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+c$
$\therefore \ I=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+c$ . Which is the required solution.
View full question & answer→Question 301 Mark
Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
AnswerGiven integrand is; $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
Let $I= \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$
Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$
$\Rightarrow \sqrt{x}=\cos \theta \text { or } \theta=\cos ^{-1} \sqrt{x}$
$\Rightarrow \mathrm{I}=\int \sqrt{\frac{1-\sqrt{\cos ^{2} \theta}}{1+\sqrt{\cos ^{2} \theta}}}(-2 \sin \theta \cos \theta) \mathrm{d} \theta$
$=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta$
$=\int-\sqrt{\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}}(2 \sin \theta \cos \theta) d \theta$
= $\int-\sqrt{\frac{\sin ^{2}\left(\frac{\theta}{2}\right)}{\cos ^{2}\left(\frac{\theta}{2}\right)}}\left(2 \sin 2 \frac{\theta}{2} \cos 2 \frac{\theta}{2}\right) d \theta$
= $\int-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot(2) \cdot\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) \cdot\left(2 \cos ^{2}\left(\frac{\theta}{2}\right)-1\right) d \theta$
$\Rightarrow \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\int-4 \cdot\left[\sin ^{2}\left(\frac{\theta}{2}\right)\right]\left(2 \cos ^{2}\left(\frac{\theta}{2}\right)-1\right) d \theta$
$=\int-4 .\left\{\left[2 . \sin ^{2}\left(\frac{\theta}{2}\right) \cos ^{2}\left(\frac{\theta}{2}\right)\right]-\sin ^{2}\left(\frac{\theta}{2}\right)\right\} \mathrm{d} \theta$
$=\int-2 \cdot\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)^{2} d \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) d \theta$
$=-2 . \int \sin ^{2} \theta \mathrm{d} \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta$
$=-2 . \int \sin ^{2} \theta \mathrm{d} \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta$
= $-2 \cdot \int \frac{1-\cos 2 \theta}{2} d \theta+4 \int \frac{1-\cos \theta}{2} d \theta$
= $-2\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]+4\left[\frac{\theta}{2}-\frac{\sin \theta}{2}\right]+C$
= $-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C$
= $\theta+\frac{2 \cdot \sin \theta \cdot \cos \theta}{2}-2 \sin \theta+c$
= $\theta+\frac{2 \cdot \sqrt{1-\cos ^{2} \theta} \cdot \cos \theta}{2}-2 \sqrt{1-\cos ^{2} \theta}+C$
= $\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C$
= $\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}-2 \sqrt{1-x}+C$
$\Rightarrow \mathrm{I}=\cos ^{-1} \sqrt{\mathrm{x}}+\sqrt{\mathrm{x}-\mathrm{x}^{2}}-2 \sqrt{1-\mathrm{x}}+\mathrm{C}$
View full question & answer→Question 311 Mark
Integrate the function $\int {\frac{{1}}{{\sqrt {{{\sin }^3}x\sin (x + \alpha )} }}} $
Answer I = $\int {\frac{{dx}}{{\sqrt {{{\sin }^3}x.\sin (x + \alpha )} }}} $ $=\int {\frac{{dx}}{{\sqrt {{{\sin }^4}x.\frac{{\sin (x + \alpha )}}{{\sin x}}} }}} $
$=\int {\frac{{dx}}{{{{\sin }^2}x\sqrt {\frac{{\sin (x + \alpha )}}{{\sin x}}} }} = \int {\frac{{\ cose{c^2}dx}}{{\sqrt {\frac{{\sin (x + \alpha )}}{{\sin x}}} }}} } $
$=\int {\frac{{\ cose{c^2}xdx}}{{\sqrt {\frac{{\sin x.\cos \alpha + \cos x.\sin \alpha }}{{\sin x}}} }}} $
$=\int {\frac{{\cos e{c^2}2dx}}{{\sqrt {\cos \alpha + \cot x.\sin \alpha } }}} $
Put $\cos \alpha + \cot x.\sin \alpha = t$
$0 - \ cose{c^2}x.\sin \alpha dx = dt$
$\ cose{c^2}xdx = \frac{{ - dt}}{{\sin \alpha }}$
$\therefore$$ I= \frac{{ - 1}}{{\sin \alpha }}\int {\frac{{dt}}{{\sqrt t }} = \frac{{ - 1}}{{\sin \alpha }}} \times \frac{{{t^{1/2}}}}{{1/2}} + c$
$ = \frac{{ - 2\sqrt t }}{{\sin \alpha }} + c$
$ = \frac{{ - 2\sqrt {\cos \alpha + \cot x\sin \alpha } }}{{\sin \alpha }} + c$
View full question & answer→Question 321 Mark
Integrate the function $f^{\prime}(a x+b)[f(a x+b)]^{n}$
AnswerLet f(ax + b) = t $\Rightarrow$ a.f'(ax + b)dx = dt
$\Rightarrow \int \mathrm{f}^{\prime}(\mathrm{ax}+\mathrm{b})\left[\mathrm{f}(\mathrm{ax}+\mathrm{b})^{\mathrm{n}}\right]=\int \mathrm{t}^{\mathrm{n}}\left(\frac{\mathrm{dt}}{\mathrm{a}}\right)$
$=\frac{1}{a} \int t^{n} d t$
= $\frac{1}{a} \cdot \frac{t^{n+1}}{n+1}+c$
= $\frac{1}{a} \cdot \frac{(f(a x+b))^{n+1}}{n+1}+C$
= $\frac{1}{a(n+1)} \cdot(f(a x+b))^{n+1}+c$
View full question & answer→Question 331 Mark
Integrate the function $e^{3 \log x}\left(x^{4}+1\right)^{-1}$
AnswerLet $\mathrm{I}=\mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1}$
$=e^{\log x^{3}}\left(x^{4}+1\right)^{-1}=\frac{x^{3}}{x^{4}+1}$
Let $x^4 = t \Rightarrow 4x^3 dx = dt \Rightarrow x^3 dx = \frac{dt}{4}$
$\Rightarrow \int e^{3 \log x}\left(x^{4}+1\right)^{-1}=\int \frac{x^{3}}{x^{4}+1} d x$
= $\int \frac{1}{t+1} \cdot \frac{d t}{4}$
= $\frac{1}{4} \cdot \int \frac{1}{t+1} \cdot d t$
= $\frac{1}{4} \log (t+1)+C$
$\Rightarrow I=\frac{1}{4} \log \left(x^{4}+1\right)+c$
View full question & answer→Question 341 Mark
Integrate the function $\int {{{\cos }^3}x{e^{\log \sin x}}}$
Answer$I=\int {{{\cos }^3}x.{e^{\log \sin x}}} dx$ $\because {e^{\log \theta }} = \theta $
$\therefore {e^{\log \sin x}} = \sin x$
$ I= \int {{{\cos }^3}} x.\sin xdx$
Put cos x = t
$ - \sin x\,dx = dt$
$\sin x\,dx = - dt$
$I = \int { - {t^3}dt} $
$ = - \frac{{{t^4}}}{4} + c = - \frac{{{{\cos }^4}x}}{4} + c$
View full question & answer→Question 351 Mark
Integrate the function $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
AnswerGiven: $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
Let $I=\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
Using partial fraction:
Let $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ .....(i)
$\Rightarrow \frac{1}{(x+1)\left(x^{2}+9\right)}$ = $\frac{(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)}{(x+1)\left(x^{2}+9\right)}$
$\Rightarrow 1 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1)$
$\Rightarrow 1 = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D$
$\Rightarrow 1 = (A + C)x^3 +(B + D)x^2 +(4A + C)x + (4B + D)$
Equating the coefficients of $x, x^2, x^3$ and constant value. We get:
A + C = 0 $\Rightarrow$ C = -A
B + D = 0 $\Rightarrow$ B = -D
4A + C = 0 $\Rightarrow$ 4A = -C $\Rightarrow$ 4A = A $\Rightarrow$ 3A = 0 $\Rightarrow$ A = 0 $\Rightarrow$ C = 0
4B + D = 1 $\Rightarrow$ 4B - B = 1 $\Rightarrow$ B = $\frac{1}{3}$ $\Rightarrow$ D = $\frac{-1}{3}$
Put these values in equation (i)
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=$ $\frac{(0) x+\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{(0) x+\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+4\right)} d x$
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1^{2}\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+2^{2}\right)} d x$
= $\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$
$\Rightarrow \mathrm{I}=\frac{1}{3} \cdot \tan ^{-1} \mathrm{x}-\frac{1}{6} \tan ^{-1} \frac{\mathrm{x}}{2}+\mathrm{C}$
View full question & answer→Question 361 Mark
Integrate the function $\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}$
AnswerLet, $I=\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}$
Let $e^x = t \Rightarrow e^x dx = dt$
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)\left(2+\mathrm{e}^{\mathrm{x}}\right)} \mathrm{d} \mathrm{x}=\int \frac{1}{(1+\mathrm{t})(2+\mathrm{t})} \mathrm{dt}$
= $\int\left[\frac{1}{(1+t)}-\frac{1}{(2+t)}\right] d t$
= $\int\left[\frac{1}{(1+t)}\right] d t-\int\left[\frac{1}{(2+t)}\right] d t$
= $\log |(1+t)|-\log |(2+t)|+c$
= $\log \left|\frac{1+\mathfrak{t}}{2+\mathfrak{t}}\right|+{C}$
$\Rightarrow \mathrm{I}=\log \left|\frac{1+\mathrm{e}^{\mathrm{x}}}{2+\mathrm{e}^{\mathrm{x}}}\right|+\mathrm{C}$
View full question & answer→Question 371 Mark
Integrate the function $\frac{x^{3}}{\sqrt{1-x^{8}}}$
AnswerLet $I=\frac{x^{3}}{\sqrt{1-x^{8}}}$
Now, let $x^4 = t \Rightarrow 4x^3 dx = dt$
And $x^3 dx = \frac{dt}{4}$
$\Rightarrow \int \frac{\mathrm{x}^{3}}{\sqrt{1-\mathrm{x}^{8}}} \mathrm{dx}=\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}}\left(\frac{\mathrm{dt}}{4}\right)$
$=\frac{1}{4} \int \frac{1}{\sqrt{1^{2}-t^{2}}} \cdot d t$
= $\frac{1}{4} \sin ^{-1} t+C$
$\Rightarrow \mathrm{I}=\frac{1}{4} \sin ^{-1}\left(\mathrm{x}^{4}\right)+\mathrm{C}$
View full question & answer→Question 381 Mark
Integrate the function $\frac{1}{\cos (x+a) \cos (x+b)}$
AnswerGiven function is: $\frac{1}{\cos (x+a) \cos (x+b)}$
Let $I=\frac{1}{\cos (x+a) \cos (x+b)}$
Multiply and divide by sin (a - b), we get
$I=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b+x-x)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}\right)$
As, {sin (A - B) = sin A cos B - cos A sin B}
$\Rightarrow$ I = $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)}{\cos (x+a) \cos (x+b)}-\frac{\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot[\tan (x+a)-\tan (x+b)]$
$\Rightarrow \int \frac{1}{\cos (x+a) \cos (x+b)} d x$ = $\int \frac{1}{\sin (a-b)} \cdot[\tan (x+a)-\tan (x+b)] d x$
= $\frac{1}{\sin (a-b)}\left\{\int \tan (x+a) d x-\int \tan (x+b) d x\right\}$
= $\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|-(-\log |\cos (x+b)|)]$
= $\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b)|]$
$\Rightarrow \mathrm{I}=\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \cdot \log \left|\frac{\cos (\mathrm{x}+\mathrm{b})}{\cos (\mathrm{x}+\mathrm{a})}\right|+\mathrm{c}$
View full question & answer→Question 391 Mark
Integrate the function $\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$
Answer$\int {\frac{{{{\left( {{{\sin }^4}x} \right)}^2} - {{\left( {{{\cos }^4}x} \right)}^2}}}{{1 - 2{{\sin }^2}x.{{\cos }^2}x}}} dx$ =$\int {\frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}{{1 - 2{{\sin }^2}x.{{\cos }^2}x}}} dx$
=$\int {\frac{{\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}{{\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)}}} dx$
=$\int { \left( {{{\sin }^2}x - {{\cos }^2}x} \right)} dx$
=$\int { - \cos 2xdx} $
=$- \frac{{\sin 2x}}{2} + c$
View full question & answer→Question 401 Mark
Integrate the function $\frac{1}{x-x^{3}}$
AnswerGiven $\frac{1}{x-x^{3}}$
Let $I=\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1+x)(1-x)}$
Using partial fraction:
Let $\frac{1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x}$ ...(i)
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow 1 = A - Ax^2 + Bx - Bx^2 + Cx + Cx^2$
$\Rightarrow 1 = A + (B + C)x + (-A - B + C)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
A = 1,
B + C = 0 $\Rightarrow$ B = -C
-A - B + C = 0
⇒ -1 - (- C) +C = 0
⇒ 2C = 1 ⇒ C = $\frac{1}{2}$
So, B = $-\frac{1}{2}$
Put these values in equation (i)
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{1}{x}+\frac{-\left(\frac{1}{2}\right)}{1+x}+\frac{\left(\frac{1}{2}\right)}{1-x}$
$\Rightarrow \int \frac{1}{x(1+x)(1-x)} dx$ = $\int \frac{1}{x} d x-\frac{1}{2} \int \frac{1}{1+x} d x+\frac{1}{2} \int \frac{1}{1-x} d$
= $\log |\mathrm{x}|-\frac{1}{2} \log |1+\mathrm{x}|+\frac{1}{2} \log |1-\mathrm{x}|$
= $\log |x|-\log \left|(1+x)^{\frac{1}{2}}\right|+\log \left|(1-x)^{\frac{1}{2}}\right|$
= $\log \left|\frac{x}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}\right|+C$
= $\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{(1+x)(1-x)^{\frac{1}{2}}}\right|+C$
= $\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{\left(1-x^{2}\right)^{\frac{1}{2}}}\right|+C$
= $\log \left|\left(\frac{x^{2}}{1-x^{2}}\right)^{\frac{1}{2}}\right|+C$
$\Rightarrow \mathrm{I}=\frac{1}{2} \log \left|\frac{\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right|+\mathrm{C}$
View full question & answer→Question 411 Mark
The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is
AnswerGiven: $\int_{\frac{1}{3}}^{1}\left(\frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\right) d x$
Let $I=\int_{\frac{1}{3}}^{1}\left(\frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\right) d x$
Now, let x = sin $\theta$ $\Rightarrow$ dx = cos $\theta$ d$\theta$
Now, when $\mathrm{x}=\frac{1}{3}, \theta=\sin ^{-1}\left(\frac{1}{3}\right)$ and when $x=1, \theta=\frac{\pi }{2}$
$\Rightarrow \mathrm{I}=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{\left(\sin \theta-\sin ^{3} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta \mathrm{d} \theta$
$=\int_{\sin ^{-1}\left (\frac{1}{3} \right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}} )\left(1-\sin ^{2} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}}\left(\cos ^{2} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}}(\cos \theta)^{\frac{2}{3}}}{\sin ^{2} \theta \cdot \sin ^{2} \theta}\right) \cos \theta d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\cos \theta)^{\frac{2}{3}+1}}{(\sin \theta)^{2-\frac{1}{3}}}\right) \cdot \frac{1}{\sin ^{2} \theta} d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\cos \theta)^{\frac{5}{3}}}{(\sin \theta)^{\frac{5}{3}}}\right) \cdot cosec ^{2} \theta d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left((\cot \theta)^{\frac{5}{3}}\right) \cdot cosec ^{2} \theta d \theta$
$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left((\cot \theta)^{\frac{5}{3}}\right) \cdot \ cosec ^{2} \theta d \theta$
put, $cot\theta=t,~then~-cosec^2\theta d\theta=dt$
When, $\theta=\sin ^{-1}\left(\frac{1}{3}\right)$, $t=2 \sqrt{2}$ and when $\theta=\frac{\pi}{2}, \mathrm{t}=0$
$\therefore$ I $=\int_{2 \sqrt{2}}^{0}-(t)^{\frac{5}{3}} \cdot d t$
$=-\left[\frac{(t)^{\frac{5}{3}+1}}{\frac{5}{3}+1}\right]_{2 \sqrt{2}}^{0}$
$=-\left[\frac{(t)^{\frac{8}{3}}}{\frac{8}{3}}\right]_{2 \sqrt{2}}^{0}$
$=-\frac{3}{8}\left[(0)^{\frac{8}{3}}-(2 \sqrt{2})^{\frac{8}{3}}\right]$
$=-\frac{3}{8}\left[-(\sqrt{8})^{\frac{8}{3}}\right]$
$=\frac{3}{8}\left[(8)^{\frac{4}{3}}\right]$
$=\frac{3}{8}[16]$
= 6
View full question & answer→Question 421 Mark
If $f(x)=\int_{0}^{x} t \sin t d t$, then f'(x) is
AnswerGiven: $f(x)=\int_{0}^{x} t \sin t d t$
Applying product rule,
$\Rightarrow \int \mathrm{u} . \mathrm{v} \mathrm{d} \mathrm{x}=\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
So, $f(x)=[t]_{0}^{x} \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \cdot \int \sin t d t\right\} d t$
$=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t$
$=[-t(\cos t)+\sin t]_{0}^{x}$
= -x cos x + sin x - 0
⇒ f(x) = -x cos x + sinx
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\left[\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}+\cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}\right]$
⇒ f'(x) = -[{x(-sinx )} + cosx ] + cosx
= x sin x - cos x + cos x
= x sin x
View full question & answer→Question 431 Mark
$\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ equals
AnswerLet $I=\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$
Taking 9 common from Denominator in I
$\Rightarrow I=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\frac{4}{9}+x^{2}}=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\left(\frac{2}{3}\right)^{2}+x^{2}}$ [$\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$]
$\Rightarrow \mathrm{I}=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{\mathrm{x}}{\frac{2}{3}}\right]_{0}^{\frac{2}{3}}$ = $\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 x}{2}\right]_{0}^{\frac{2}{3}}$
$\Rightarrow I=\frac{1}{6}\left[\tan ^{-1} \frac{3}{2} \times \frac{2}{3}-\tan ^{-1} 0\right]$ = $\frac{1}{6}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]$
$\Rightarrow I=\frac{1}{6} \times\left(\frac{\pi}{4}-0\right)=\frac{\pi}{ 24}$
$\therefore \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}=\frac{\pi}{24}$
View full question & answer→Question 441 Mark
$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$ equals
AnswerLet $I=\int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1}$
$\Rightarrow I=\int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1}$
$\Rightarrow I=\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}}$ = $\left[\tan ^{-1} \sqrt{3}-\tan ^{-1} 1\right]$ = $\frac{\pi}{3}-\frac{\pi}{4}$ ...[$\because~\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$]
$\Rightarrow I=\frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}$
$\therefore \int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1}=\frac{\pi}{12}$
View full question & answer→Question 451 Mark
$\int \sqrt{x^{2}-8 x+7} d x$ is equal to
Answer$I=\int \sqrt{x^{2}-8 x+7} d x$
$=\int \sqrt{\left(x^{2}-8 x+16\right)-9} d x$
$=\int \sqrt{(x-4)^{2}-(3)^{2}} d x$
We know that
$\Rightarrow \int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
Therefore,
$\Rightarrow I=\frac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log |(x-4)+\sqrt{x^{2}-8 x+7}|+C$
View full question & answer→Question 461 Mark
$\int \sqrt{1+x^{2}} d x$ is equals to
AnswerWe know that,
$\Rightarrow \int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$
Therefore,
$\Rightarrow \int \sqrt{1+x^{2}} d x=\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log |x+\sqrt{1+x^{2}}|+C$
View full question & answer→Question 471 Mark
Integrate the function $x \cos^{-1} x$
AnswerLet I = $\int$$x .cos^{-1} x$
Now, integrating by parts, we get,
$I=\cos ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int x d x\right\} d x$
$=$ $\cos ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x$
$=$$\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$
= $\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int\left\{\sqrt{1-x^{2}}+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\right\} d x$
= $\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \sqrt{1-x^{2}} d x-\frac{1}{2} \int\left(\frac{-1}{\sqrt{1-x^{2}}}\right) d x$
= $\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} I_{1}-\frac{1}{2} \cos ^{-1} x$ ...(i)
Now, $I_{1}=\int \sqrt{1-x^{2}} d x$
$I_{1}=x \sqrt{1-x^{2}}-\int \frac{d}{d x} \sqrt{1-x^{2}} \int d x$
$I_{1}=x \sqrt{1-x^{2}}-\int \frac{-2 x}{2 \sqrt{1-x^{2}}} x \cdot d x$
= $x \sqrt{1-x^{2}}-\int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x$
= $x \sqrt{1-x^{2}}-\int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$
= $x \sqrt{1-x^{2}}-\left\{\int \sqrt{1-x^{2}} d x+\int \frac{-d x}{\sqrt{1-x^{2}}}\right\}$
$\therefore I_{1}=x \sqrt{1-x^{2}}-\left\{I_{1}+\cos ^{-1} x\right\}$
$\Rightarrow 2I_{1}=x \sqrt{1-x^{2}}-\cos ^{-1} x$
$\Rightarrow I_{1}=\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x$
Now, substituting in (i), we get,
$I=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2}\left(\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x\right)-\frac{1}{2} \cos ^{-1} x$
= $\frac{\left(2 x^{2}-1\right)}{4} \cos ^{-1} x-\frac{x}{4} \sqrt{1-x^{2}}+C$
View full question & answer→Question 481 Mark
Integrate the function $x \tan^{-1}x$
AnswerLet $I = \int {x{{\tan }^{ - 1}}x} dx$$ = \int {\left( {{{\tan }^{ - 1}}x} \right).x} dx$
$= \left( {{{\tan }^{ - 1}}x} \right).\frac{{{x^2}}}{2} - \int {\frac{1}{{1 + {x^2}}}.\frac{{{x^2}}}{2}dx} $
$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} $
$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2} + 1 - 1}}{{{x^2} + 1}}dx}$
$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {1 - \frac{1}{{{x^2} + 1}}} \right)dx}$
$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\left( {x - {{\tan }^{ - 1}}x} \right) + c$
$= \frac{1}{2}\left[ {{x^2}{{\tan }^{ - 1}}x - x + {{\tan }^{ - 1}}x} \right] + c$
$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{x}{2} + \frac{1}{2}{\tan ^{ - 1}}x + c$
View full question & answer→Question 491 Mark
Integrate the function $x \sin^{-1} x$
AnswerLet $I = x \sin^{-1}x$
Now, integrating by parts, we get,
$I=\sin ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right\} d x$
= $\sin ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x$
= $\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x$
= $\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\frac{1-x^{2}}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x$
= $\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\sqrt{1-x^{2}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x$
= $\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\left\{\int \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}} d x\right\}$
= $\frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C$
= $\frac{1}{4}\left(2 x^{2}-1\right) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^{2}}+C$
View full question & answer→Question 501 Mark
Integrate the function $x^2 \log x$
Answer$\int {{x^2}\log xdx} $$= \int {\left( {\log x} \right){x^2}dx} $
$= \log x\int {{x^2}dx - \int {\left( {\frac{d}{{dx}}\log x\int {{x^2}dx} } \right)dx} } $
[Applying product rule]
$= \left( {\log x} \right)\frac{{{x^3}}}{3} - \int {\frac{1}{x}.\frac{{{x^3}}}{3}dx} $
$= \frac{{{x^3}}}{3}\log x - \frac{1}{3}\int{x^2}dx$
$= \frac{{{x^3}}}{3}\log x - \frac{1}{3}\frac{{{x^3}}}{3} + c$
$= \frac{{{x^3}}}{3}\log x - \frac{{{x^3}}}{9} + c$
View full question & answer→Question 511 Mark
Integrate the function x log 2x
Answer$\int {x\log 2xdx} $ $ = \int {\left( {\log 2x} \right)xdx} $
$= \left( {\log 2x} \right)\int {xdx - \int {\left[ {\frac{d}{{dx}}\log 2x\int {xdx} } \right]dx} } $
[Applying product rule]
$= \left( {\log 2x} \right)\frac{{{x^2}}}{2} - \int {\frac{1}{{2x}}.2.\frac{{{x^2}}}{2}dx} $
$ = \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\int {xdx} $
$= \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\frac{{{x^2}}}{2} + c$
$= \frac{{{x^2}}}{2}\log 2x - \frac{{{x^2}}}{4} + c$
View full question & answer→Question 521 Mark
Integrate the function x log x
AnswerLet $ I = x log x$
Now, integrating by parts, we get,
Taking, Logarithmic function as first function and algebraic function as second function,
$I=\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x$
= $\log x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$
= $\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x$
= $\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C$
View full question & answer→Question 531 Mark
Integrate the function $x^2e^x$
Answer$\int {{x^2}{e^x}} dx$$ = {x^2}\int {{e^x}dx - \int {\left[ {\frac{d}{{dx}}{x^2}\int {{e^x}dx} } \right]} dx} $
[Applying product rule]
$ = {x^2}{e^x} - \int {2x{e^x}dx} $
$ = {x^2}{e^x} - 2\int {x{e^x}dx} $
$= {x^2}{e^x} - 2\left[ {{xe^x}dx - \int {\left\{ {\frac{d}{{dx}}x\int {{e^x}dx} } \right\}dx} } \right]$
[Again applying product rule]
$= {x^2}{e^x} - 2\left( {x{e^x} - \int {1.{e^x}dx} } \right)$
$= {x^2}{e^x} - 2\left( {x{e^x} - \int {{e^x}dx} } \right)$
$ = {x^2}{e^x} - 2x{e^x} + 2\int {{e^x}dx} $
$= x^2e^x - 2xe^x + 2e^x + c$
$= e^x(x^2 - 2x + 2) + c$
View full question & answer→Question 541 Mark
$\int e^{x} \sec x(1+\tan x) d x$ equals
Answer$\int e^{x} \sec x(1+\tan x) d x$
Let
$\mathrm{I}=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x$ ...(i)
Also, let secx = f(x)
$\Rightarrow$ secxtanx = f’(x)
We know that $\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+C$
Thus (i) gives, $I = e^xsec\ x + C$
View full question & answer→Question 551 Mark
$\int x^{2} e^{x^{3}} d x$ equals
AnswerLet $I=\int x^{2} e^{x^{3}} d x$
Also, let $x^3 = t, \Rightarrow 3x^2dx = dt$
Thus,
$\Rightarrow I=\frac{1}{3} \int e^{t} d t$
$=\frac{1}{3}\left(e^{t}\right)+C$
$=\frac{1}{3}\left(e^{x^{3}}\right)+C$
View full question & answer→Question 561 Mark
Integrate the function ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$
AnswerPutting $x = \tan \theta $
$ \Rightarrow dx = {\sec ^2}\theta d\theta $
$\therefore \int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} $
$= \int {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right).{{\sec }^2}\theta d\theta }$
$= \int {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right).{{\sec }^2}\theta d\theta } $
$= \int {2\theta {{\sec }^2}\theta d\theta }$
$ = 2\int {\theta {{\sec }^2}\theta d\theta } $ ...[Applying product rule]
$= 2\left[ {\theta .\tan \theta - \int {1.\tan \theta d\theta } } \right]$
$= 2\left[ {\theta .\tan \theta - \int {\tan \theta d\theta } } \right]$
$ = 2\left[ {\theta \tan \theta - \log \sec \theta } \right] + c$
$= 2\left[ {{{\tan }^{ - 1}}x.x - \log \sqrt {1 + {x^2}} } \right] + c$
$= 2\left[ {x{{\tan }^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right)} \right] + c$
$= 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + c$
View full question & answer→Question 571 Mark
Integrate the function $e^{2x}$ sin $x$
AnswerLet $I$ = $e^{2x}$ sin $x$
Integrating by parts, we get,
$I=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x$
= $\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$
= $\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$
Again, integrating by parts, we get,
$I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \cos x\right) \int e^{2 x} d x\right\} d x\right]$
= $\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \cdot \frac{e^{2 x}}{2} d x\right]$
= $\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I$
$\Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow \frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]$
View full question & answer→Question 581 Mark
Integrate the function $\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}$
AnswerLet $I = \frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx$
$= \int {\frac{{\left( {x - 1} \right) - 2}}{{{{\left( {x - 1} \right)}^3}}}{e^x}} dx$
$= \int {{e^x}\left[ {\frac{{\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^3}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\Rightarrow I = \int {{e^x}\left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} } \right]$
It is in the form of ${\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} }$ since here $f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}$ and $f'\left( x \right) = \frac{d}{{dx}}\left\{ {{{\left( {x - 1} \right)}^{ - 2}}} \right\}$
$ = - 2{\left( {x - 1} \right)^{ - 3}}$
$ = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}$.
$ \Rightarrow I = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$
View full question & answer→Question 591 Mark
Integrate the function x sin 3x
Answer$\int {x\sin 3xdx} $ $= x\int {\sin 3xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin 3x} dx} \right)dx} } $
[Applying product rule]
$= x\left( {\frac{{ - \cos 3x}}{3}} \right) - \int {1\left( {\frac{{ - \cos 3x}}{3}} \right)dx + c}$
$ = \frac{{ - 1}}{3}x\cos 3x + \frac{1}{3}\int {\cos 3xdx + c} $
$= \frac{{ - 1}}{3}x\cos 3x + \frac{1}{3}\frac{{\sin 3x}}{3} + c$
$= \frac{{ - 1}}{3}x\cos 3x + \frac{1}{9}\sin 3x + c$
$= - \frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + c$
View full question & answer→Question 601 Mark
Integrate the function ${e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$
AnswerLet $I = {e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx$ $\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} } \right]$
It is in the form of ${\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} }$, Here $f\left( x \right) = \frac{1}{x} = {x^{ - 1}}$ and $f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}$
$ \Rightarrow I = {e^x}\frac{1}{x} + c$
$= \frac{{{e^x}}}{x} + c$ ...$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx = {e^x}f\left( x \right) + c} \right]$
View full question & answer→Question 611 Mark
Integrate the function ${e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right)$
AnswerLet $I = \int {{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}dx} $ $ = \int {{e^x}.\frac{{1 + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}dx} $
$ = \int {{e^x}\left[ {\frac{1}{{2{{\cos }^2}\frac{x}{2}}} + \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right]dx} $
$= \int {{e^x}\left( {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right)dx}$
$= \int {{e^x}\left( {\tan \frac{x}{2} + \frac{1}{2}{{\sec }^2}\frac{x}{2}} \right)} dx$
$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx} \right]$
It is in the form of $\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx$ since here $f\left( x \right) = \tan \frac{x}{2}$ and $f'\left( x \right) = \frac{1}{2}{\sec ^2}\frac{x}{2}$
$ = {e^x}\tan \frac{x}{2} + c$
$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$
View full question & answer→Question 621 Mark
Integrate the function $\frac{x e^{x}}{(1+x)^{2}}$
Answer$I=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{\frac{x}{(1+x)^{2}}\right\} d x$
= $\int e^{x}\left\{\frac{1+x-1}{(1+x)^{2}}\right\} d x$
= $\int e^{x}\left\{\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right\} d x$
Now,
Let $\frac{1}{1+x}=f(x) \Rightarrow f^{\prime}(x)=-\frac{1}{(1+x)^{2}}$
We know that,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C$
Thus,
$\int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+C$
View full question & answer→Question 631 Mark
Integrate the function $e^x (\sin x + \cos x)$
Answer$I=\int e^{x}(\sin x+\cos x) d x$
Now,
Let sin x = f(x) $\Rightarrow$ f'(x) = cos x
We know that,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Thus,
$\int e^{x}(\sin x+\cos x) d x=e^{x} \sin x+C$
View full question & answer→Question 641 Mark
Integrate the function $(x^2 + 1)$ log $x$
Answer$\int {\left( {{x^2} + 1} \right)\log xdx} $
$ = \int {\left( {\log x} \right)\left( {{x^2} + 1} \right)dx} $
[Applying product rule]
$= \log x\left( {\frac{{{x^3}}}{3} + x} \right) - \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \int {\left( {\frac{{{x^2}}}{3} + 1} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {{x^2}dx - \int {1dx} } $
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\frac{{{x^3}}}{3} - x + c$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{{{x^3}}}{9} - x + c$
View full question & answer→Question 651 Mark
Integrate the function x $(\log x)^2$
AnswerLet $I=x(\log x)^{2}$
Integrating by parts, we get,
$I=\left[(\log x)^2 \int x d x-\int\left\{\left(\frac{d}{d x} (\log x)^2 \right) \int x d x\right\} d x\right]$
= $\left[\frac{x^{2}}{2}(\log x)^2-\int \frac{2 \log x }{x} \cdot \frac{x^{2}}{2} d x\right]$
= $\frac{x^{2}}{2}(\log x)^{2}- \int x \log x \cdot d x$
= $\frac{x^{2}}{2}(\log x)^{2} - [ log x ∫ x dx -∫( \frac {d}{dx} (log x)∫ x dx) dx]$
= $\frac{x^{2}}{2}(\log x)^{2} - [ \frac {x²}{2}\log x - ∫(\frac1x \cdot\frac{x^2}{2} ) dx$
= $\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C$
View full question & answer→Question 661 Mark
Integrate the function $\tan^{-1}x$
AnswerLet $I = \tan^{-1}xdx$
$= \int {\left( {{{\tan }^{ - 1}}x} \right).1} dx$
$ = {\tan ^{ - 1}}x.x - \int {\frac{1}{{1 + {x^2}}}x.dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {\left( {1 + {x^2}} \right)} \right| + c$
$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \log \left| {f\left( x \right)} \right|} } \right]$
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + c$
View full question & answer→Question 671 Mark
Integrate the function $x \sec^2 x$
AnswerLet $I = x \sec^2x$
Now, integrating by parts, we get,
$I=x \int \sec ^{2} x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sec ^{2} x d x\right\} d x$
= $x \tan x-\int (1) (\tan x) d x$
= $x \tan x+\log |\cos x|+c$
View full question & answer→Question 681 Mark
Integrate the function: $\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$
AnswerLet $I=\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$
$I=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$
Now, integrating by parts, we get,
$I = \frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right]$
= $-\frac{1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right]$
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x\right]$
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+C$
= $-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+C$
View full question & answer→Question 691 Mark
Integrate the function $(\sin^{-1}x)^2$
AnswerPutting $x = \sin \theta \Rightarrow dx = \cos \theta d\theta $$\therefore \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx$
$= \int {{\theta ^2}\cos \theta } d\theta $
[Applying product rule]
$ = {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta } $
$= {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta }$
[Again applying product rule]
$= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]$
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta d\theta } $
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\sin \theta + c$
$= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c$
View full question & answer→Question 701 Mark
Integrate the function x sin x
Answer$\int {x\sin x} dx$ $= x\int {\sin x} dx - \int {\left( {\frac{d}{{dx}}x\int {\sin xdx} } \right)dx} $
[Applying product rule]
$ = x\left( { - \cos x} \right) - \int {1\left( { - \cos x} \right)dx} $
$ = -x\cos x - \int { - \cos xdx} $
$= - x\cos x + \int {\cos xdx} $
= -x cos x + sin x + c
View full question & answer→Question 711 Mark
$\int \frac{d x}{x\left(x^{2}+1\right)}$ equals
AnswerLet $\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$
$1 = A(X^2 + 1) + (Bx + C)x = Ax^2 + A + Bx^2 + Cx = (A + B)x^2+Cx + A$
Equating the coefficients of $x^2, x$ and constant term, we get,
A + B = 0
C = 0
A = 1
On solving these equations, we get,
A = 1, B = -1 and C = 0
Therefore, $\frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
$\int \frac{1}{x\left(x^{2}+1\right)}=\int\left\{\frac{1}{x}+\frac{-x}{x^{2}+1}\right\} d x$
= $\log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C$
View full question & answer→Question 721 Mark
$\int \frac{x d x}{(x-1)(x-2)}$ equals
AnswerLet $\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$
x = A(x - 2) + B(x - 1) …(i)
Substituting x = 1 and 2 in (i), we get,
A = -1 and B = 2
Therefore, $\frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$
$\int \frac{x}{(x-1)(x-2)} d x=\int\left\{-\frac{1}{(x-1)}+\frac{2}{(x-2)}\right\} d x$
= $-\log |x-1|+2 \log |x-2|+c$
= $\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
View full question & answer→Question 731 Mark
$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$ equals
Answer$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}=\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x\right)}}$
$=\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64}\right)}}$
$=\int \frac{d x}{\sqrt{-4\left[\left(x-\frac{9}{8}\right)^{2}-\left(\frac{9}{8}\right)^{2}\right.}]}$
$=\frac{1}{2}\left[\sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)\right]+C$
$=\frac{1}{2}\left[\sin ^{-1}\left(\frac{8 x-9}{9}\right)\right]+C$
View full question & answer→Question 741 Mark
$\int \frac{d x}{x^{2}+2 x+2}$ equals
Answer$\int \frac{d x}{x^{2}+2 x+2}=\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}$
$= \int \frac{1}{(x+1)^{2}+(1)^{2}} d x$
$=\left[\tan ^{-1}(x+1)\right]+C$
View full question & answer→Question 751 Mark
$\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals
AnswerLet I = $\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$
$=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x~~~$
$=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$
$= \int \sec ^{2} x d x+\int cosec ^{2} x d x$
$$= tanx - cotx + C
View full question & answer→Question 761 Mark
$\int \frac{10 x^{9}+10^{x} \log _{e} 10 d x}{x^{10}+10^{x}}$, equals
AnswerLet $x^{10}+10^x=t$
$\Rightarrow\left(10 x^9+10^x \log _e 10\right) d x=d t$
$\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$
$= \log t + C$
$= \log(x^{10} + 10^x) + C$
View full question & answer→Question 771 Mark
If $\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$ such that f(2) = 0. Then f(x) is
AnswerIt if given that $\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$
$f(x)=\int 4 x^{3}-\frac{3}{x^{4}} d x$
$\Rightarrow f(x)=4 \int x^{3} d x-3 \int\left(x^{-4}\right) d x$
$\Rightarrow f(x)=4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C$
$\Rightarrow f(x)=x^{4}+\frac{1}{x^{3}}+C$
Also, It is given that f(2) = 0
$\Rightarrow f(2)=(2)^{4}+\frac{1}{(2)^{3}}+\mathrm{C}=0$
$\Rightarrow 16+\frac{1}{8}+C=0$
$\Rightarrow \mathrm{C}=-\left(16+\frac{1}{8}\right)=-\frac{129}{8}$
Therefore, $f(x)=x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$. Which is the required solution.
View full question & answer→Question 781 Mark
The anti derivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
Answer$\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
= $\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x$
= $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$
= $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C$
View full question & answer→Question 791 Mark
The value of $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is
AnswerGiven Integral is: $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ ......(i)
As, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-\mathrm{x}\right)}{4+3 \cos \left(\frac{\pi}{2}-\mathrm{x}\right)}\right) \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos \mathrm{x}}{4+3 \sin x}\right) \mathrm{d} \mathrm{x}$ ....(ii)
Adding (i) and (ii), we get
$~~~~~2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left\{\log \left(\frac{4+3 \sin \mathrm{x}}{4+3 \cos \mathrm{x}}\right)+\left(\frac{4+3 \cos \mathrm{x}}{4+3 \sin }\right)\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \mathrm{x}}{4+3 \cos \mathrm{x}} \times \frac{4+3 \cos \mathrm{x}}{4+3 \sin x }\right) \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log 1 \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} 0 . \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=0$
View full question & answer→Question 801 Mark
The value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$ is
AnswerGiven: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$
Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$
$\Rightarrow \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\mathrm{x}^{3}\right) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{x} \cos \mathrm{x}) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\tan ^{5} \mathrm{x}\right) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1) \mathrm{dx}$
It is also known that if f(x) is an even function then,
$\Rightarrow \mathrm{I}=0+0+0+2 \cdot \int_{0}^{\frac{\pi}{2}}(1) \mathrm{d} \mathrm{x}$ ...$\left\{\int_{-a}^{a} f(x) d x=0\right\}$
$\Rightarrow \mathrm{I}=2 \cdot[\mathrm{x}]_{0}^{\frac{\pi}{2}}$
$\Rightarrow I=2 \cdot \frac{\pi}{2}$
$\Rightarrow \mathrm{I}=\pi$
View full question & answer→Question 811 Mark
Find the integral: $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}$
AnswerWe have $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}$
= $\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$ (completing the square)
Put $x-\frac{1}{5}=t$. Then dx = dt
Therefore, $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}$
= $\frac{1}{\sqrt{5}} \log |t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}|+C$
= $\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+C$
View full question & answer→Question 821 Mark
Find the integral: $\int \frac{d x}{3 x^{2}+13 x-10}$
AnswerWe write the denominator of the integrand,
$3x^2 + 13x - 10 = 3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)$
= $3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right]$ (completing the square)
Thus, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$
Put $x+\frac{13}{6}=t$. Then dx = dt
Therefore, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}}$
= $\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+\mathrm{C}_{1}$
= $\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1}$
= $\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1}$
= $\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3}$
= $\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C$ where, $\mathrm{C}=\mathrm{C}_{1}+\frac{1}{17} \log \frac{1}{3}$
View full question & answer→Question 831 Mark
Find the integral: $\int \frac{d x}{x^{2}-6 x+13}$
AnswerWe have $x^2 - 6x + 13 = x^2 - 6x + 3^2 - 3^2 + 13 = (x - 3)^2 + 4$
So, $\int \frac{d x}{x^{2}-6 x+13}=\int \frac{1}{(x-3)^{2}+2^{2}} d x$
Let x – 3 = t $\Rightarrow$ dx = dt
Therefore,
$\int \frac{d x}{x^{2}-6 x+13}=\int \frac{d t}{t^{2}+2^{2}}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+\mathrm{C}$
$=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+C$
View full question & answer→Question 841 Mark
Find the integral: $\int \frac{d x}{\sqrt{2 x-x^{2}}}$
AnswerLet I = $\int \frac{d x}{\sqrt{2 x-x^{2}}}$
= $\int \frac{d x}{\sqrt{1-(x-1)^{2}}}$
Put x - 1 = t. Then dx = dt.
Therefore, $\int \frac{d x}{\sqrt{2 x-x^{2}}}=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1}(t)+C$
$= \sin^{-1} (x-1) + C$
View full question & answer→Question 851 Mark
Find the integral: $\int \frac{d x}{x^{2}-16}$
AnswerWe have $\int \frac{d x}{x^{2}-16}=\int \frac{d x}{x^{2}-4^{2}}=\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|+C$$~~~(Using \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C)$
View full question & answer→Question 861 Mark
Find: $\int \sin ^{3} x d x$
AnswerFrom the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$, we find that
$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$
Therefore, $\int \sin ^{3} x d x=\frac{3}{4} \int \sin x d x-\frac{1}{4} \int \sin 3 x d x$
Integral = $-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x+c$
View full question & answer→Question 871 Mark
Find: $\int \sin 2 x \cos 3 x d x$
AnswerAs we know , sin x cos y = $\frac{1}{2}[\sin (x+y)+\sin (x-y)]$
Then , $\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]$
= $\frac{1}{2}\left[-\frac{1}{5} \cos 5 x+\cos x\right]+C$
= $-\frac{1}{10} \cos 5 x+\frac{1}{2} \cos x+C$
View full question & answer→Question 881 Mark
Find: $\int \cos ^{2} x d x$
Answer$\cos 2x = 2 \cos^2 x - 1$
$\cos ^{2} x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^{2} x d x=\frac{1}{2} \int(1+\cos 2 x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$
= $\frac{x}{2}+\frac{1}{4} \sin 2 x+C$
View full question & answer→Question 891 Mark
Find the integral: $\int \frac{1}{1+\tan x} d x$
Answer$\int \frac{d x}{1+\tan x}=\int \frac{\cos x d x}{\cos x+\sin x}$
= $\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x}$
= $\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
= $\frac{x}{2}+\frac{C_{1}}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$ ...(i)
Now, consider I = $\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
Put cos x + sin x = t so that (cos x - sin x) dx = dt
Therefore, $\mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{2}=\log |\cos x+\sin x|+\mathrm{C}_{2}$
Putting it in (i), we get
$\int \frac{d x}{1+\tan x}=\frac{x}{2}+\frac{\mathrm{C}_{1}}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{2}}{2}$
= $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{C_{1}}{2}+\frac{C_{2}}{2}$
= $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C},\left(\mathrm{C}=\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2}\right)$
View full question & answer→Question 901 Mark
Find the integral: $\int \frac{\sin x}{\sin (x+a)} d x$
AnswerPut x + a = t. Then dx = dt. Therefore
$\int \frac{\sin x}{\sin (x+a)} d x=\int \frac{\sin (t-a)}{\sin t} d t$
= $\int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t$
= $\cos a \int d t-\sin a \int \cot t d t$
= $(\cos a) t-(\sin a)\left[\log |\sin t|+C_{1}\right]$
= $(\cos a)(x+a)-(\sin a)\left[\log |\sin (x+a)|+C_{1}\right]$
= x cos a + a cos a - (sin a) log |sin (x + a)| - $C_1$ sin a
Hence, $\int \frac{\sin x}{\sin (x+a)} d x$ = x cos a - (sin a) log |sin (x + a)| - C,
where, $C = -C_1$ sin a + a cos a, is another arbitrary constant.
View full question & answer→Question 911 Mark
Find the integral: $\int \sin ^{3} x \cos ^{2} x d x$
AnswerWe have
$\int \sin ^{3} x \cos ^{2} x d x=\int \sin ^{2} x \cos ^{2} x(\sin x) d x$
= $\int\left(1-\cos ^{2} x\right) \cos ^{2} x(\sin x) d x$
Put t = cos x so that dt = -sin x dx
Therefore, $\int \sin ^{2} x \cos ^{2} x(\sin x) d x$ = $-\int\left(1-t^{2}\right) t^{2} d t$
= $-\int\left(t^{2}-t^{4}\right) d t=-\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)+\mathrm{C}$
= $-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+C$
View full question & answer→Question 921 Mark
Integrate the function w.r.t. x: $\frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}$
AnswerDerivative of $\tan ^{-1} x=\frac{1}{1+x^{2}}$. Thus, we use the substitution $\tan^{-1} x = t$ so that $\frac{d x}{1+x^{2}}=d t$
Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x=\int \sin t d t = -\cos t + C = -\cos(\tan^{-1}x) + C$
View full question & answer→Question 931 Mark
Integrate the function w.r.t. x: $\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}$
AnswerDerivative of $\sqrt{x}$ is $\frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}}$.
Thus, we use the substitution $\sqrt{x}=t$ so that $\frac{1}{2 \sqrt{x}} d x=d t$ giving dx = 2t dt.
Thus, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x$ = $\int \frac{2 t \tan ^{4} t \sec ^{2} t d t}{t}$ = $2 \int \tan ^{4} t \sec ^{2} t d t$
Again, we make another substitution tan t = u so that $\sec^2$ t dt = du
Therefore, $2 \int \tan ^{4} t \sec ^{2} t d t=2 \int u^{4} d u=2 \frac{u^{5}}{5}+C$
= $\frac{2}{5} \tan ^{5} t+C$ (since u = tan t)
= $\frac{2}{5} \tan ^{5} \sqrt{x}+C$ (since $t=\sqrt{x}$)
Hence, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x=\frac{2}{5} \tan ^{5} \sqrt{x}+C$
View full question & answer→Question 941 Mark
Integrate the function w.r.t. $x: 2x$ sin $(x^2 + 1)$
AnswerDerivative of $x^2 + 1$ is $2x$. Thus, we use the substitution $x^2 + 1 = t$ so that 2x dx = dt.
Therefore, $\int 2 x \sin \left(x^{2}+1\right) d x=\int \sin t d t=-\cos t+C=-\cos \left(x^{2}+1\right)+C$
View full question & answer→Question 951 Mark
Integrate the function sin mx w.r.t. x.
AnswerWe know that derivative of mx is m.
Thus, we make the substitution mx = t so that mdx = dt.
Therefore, $\int \sin m x d x=\frac{1}{m} \int \sin t d t$ $=-\frac{1}{m} \cos t+\mathrm{C}=-\frac{1}{m} \cos m x+\mathrm{C}$
View full question & answer→Question 961 Mark
Evaluate $ \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x.$
AnswerAccording to the question , $ I = \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(i)
$ \Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } ( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x ) } d x$$ \left[ \because \int _ { 0 } ^ { a } f ( x ) d x = \int _ { 0 } ^ { a } f ( a - x ) d x \right]$
$ \Rightarrow \quad I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(ii)
On adding Equations (i) and (ii) we get ,
$ 2 I = \int _ { 0 } ^ { \pi } \frac { ( x + \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$
$ \Rightarrow \quad 2 I = \pi \int _ { 0 } ^ { \pi } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
we know that,$ \int _ { 0 } ^ { 2 a } f ( x ) d x = 2 \int _ { 0 } ^ { a }f(x) dx \ , \ if \ \ f(2a - x) = f(x)$
Here, $ a^2 cos^2( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x )$
$= a^2 cos^2 x + b^2 sin^2x$
$ \therefore \quad 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
On dividing numerator and denominator by $cos^2x$ we get ,
$ 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { \sec ^ { 2 } x } { a ^ { 2 } + b ^ { 2 } \tan ^ { 2 } x } d x$
Put $tan x = t \implies sec^2 x dx = dt$
Lower limit when x = 0, then $t = tan 0 = 0$
Upper limit when x = $ \frac { \pi } { 2 },$then $ t =\tan \frac { \pi } { 2 } = \infty$
$ \therefore \quad I = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + b ^ { 2 } t ^ { 2 } }$
$ = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + ( b t ) ^ { 2 } }$ $ = \frac { \pi } { b ^ { 2 } } \int _ { 0 } ^ { \infty } \frac { d t } { \left( \frac { a } { b } \right) ^ { 2 } + t ^ { 2 } }$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \frac { b t } { a } \right] _ { 0 } ^ { \infty } \left[ \because \int \frac { d x } { a ^ { 2 } + x ^ { 2 } } = \frac { 1 } { a } \tan ^ { - 1 } \frac { x } { a } + C \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \infty - \tan ^ { - 1 } 0 \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \frac { \pi } { 2 } - 0 \right]$$ \left[ \begin{array} { c } { \because \tan ^ { - 1 } \infty = \tan ^ { - 1 } \left( \tan \frac { \pi } { 2 } \right) = \frac { \pi } { 2 } } \\ { \text { and } \tan ^ { - 1 } 0 = \tan ^ { - 1 } \left( \tan 0 ^ { \circ } \right) = 0 } \end{array} \right]$
$ \therefore I = \frac { \pi ^ { 2 } } { 2 a b }$
View full question & answer→Question 971 Mark
Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$
AnswerHere $ f (x) = | x \sin \pi x| = \left\{\begin{array}{l} {x \sin \pi x \text { for }-1 \leq x \leq 1} \\ {-x \sin \pi x \text { for } 1 \leq x \leq \frac{3}{2}} \end{array}\right.$
Therefore $\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x$
$= \int_{-1}^{1} x \sin \pi x \ d x+\int_{1}^{\frac{3}{2}}-x \sin \pi x\ d x$
$= \int_{-1}^{1} x \sin \pi x\ d x-\int_{1}^{\frac{3}{2}} x \sin \pi x \ d x$
Integrating both integrals on righthand side, we get
$\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x$
$= \left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{-1}^{1}-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{1}^{\frac{3}{2}}$
$= \frac{2}{\pi}-\left[-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right]=\frac{3}{\pi}+\frac{1}{\pi^{2}}$
View full question & answer→Question 981 Mark
Find $\int \frac{\sin 2 x \cos 2 x d x}{\sqrt{9-\cos ^{4}(2 x)}}$
AnswerLet $I=\int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^{4} 2 x}} d x$
Put $\cos^2$ (2x) = t so that 2 sin 2x cos 2x dx = -dt
Therefore,
I = $-\frac{1}{2} \int \frac{d t}{\sqrt{9-t^{2}}}$ = $-\frac{1}{2} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C}=-\frac{1}{2} \sin ^{-1}\left[\frac{1}{3} \cos ^{2} 2 x\right]+\mathrm{C}$
View full question & answer→Question 991 Mark
Find the anti derivative F of f defined by $f(x) = 4x^3 - 6$, where $F(0) = 3$
AnswerHere, $f(x) = x^4 - 6x$
Anti derivative F of f(x) is given by;
$F(x)=\int f(x) d x$
$=\int\left(4 x^{3}-6\right) d x$
$=x^{4}-6 x+C$
$\text { So, } F(x)=x^{4}-6 x+C, $ where C is constant of Integration.
Given: F(0) = 3, which gives,
3 = 0 - 6 $\times$ 0 + C
$\Rightarrow$ C = 3
Hence, the required anti derivative is the unique function F defined by $F(x) = x^4 - 6x + 3$
View full question & answer→Question 1001 Mark
Find $ \int ( \sqrt { \cot x } + \sqrt { \tan x } )$dx
AnswerLet $I = \int [ \sqrt { \cot x } + \sqrt { \tan x } ] d x$ $=\int \frac{1}{\sqrt{tanx}} + \sqrt{tanx} \ [\because cotx = \frac {1}{tanx}]$
$ = \int \sqrt { \tan x } \Bigg[1 +\frac {1}{\Big(\sqrt { \tan x }\Big)^2}\Bigg] d x$
put $ tan x = t^2 \implies sec^2 x dx = 2t dt$
$ \Rightarrow \quad d x = \frac { 2 t } { sec^2x}dt$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 +tan^2x }dt$$[ \because 1 + \tan ^ { 2 } x = \sec ^ { 2 } x]$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 + (t ^ { 2 })^2 }$$[ tan x = t^2 ]$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 + t ^ { 4 } }$
$ \therefore \quad I = \int t \left( 1 + \frac { 1 } { t ^ { 2 } } \right) \frac { 2 t } { \left( 1 + t ^ { 4 } \right) } d t [\because tanx =t^2 \implies \sqrt {tanx} = t]$
$ = 2 \int \frac { t ^ { 2 } + 1 } { t ^ { 4 } + 1 } d t$
On dividing numerator and denominator by $t^2$ , we get
$ I = 2 \int \frac { \left( 1 + \frac { 1 } { t ^ { 2 } } \right) } { \left( t ^ { 2 } + \frac { 1 } { t ^ { 2 } } \right) } d t $
$= 2 \int \frac { 1 + \frac { 1 } { t ^ { 2 } } } { t ^ { 2 } + \frac { 1 } { t ^ { 2 } } - 2 + 2 } d t$
$ = 2 \int \frac { \left( 1 + \frac { 1 } { t ^ { 2 } } \right) } { \left( t - \frac { 1 } { t } \right) ^ { 2 } + 2 } d t$
Again, put $ t-\frac { 1 } { t } = y \Rightarrow \left( 1 + \frac { 1 } { t ^ { 2 } } \right) d t = d y$
$ \therefore I = 2 \int \frac { d y } { y ^ { 2 } + ( \sqrt { 2 } ) ^ { 2 } } $
$ I = \frac { 2 } { \sqrt { 2 } } \tan ^ { - 1 } \frac { y } { \sqrt { 2 } } + C$$ \left[ \because \int \frac { d x } { \sqrt { x ^ { 2 } + a ^ { 2 } } } = \frac { 1 } { a } \tan ^ { - 1 } \left( \frac { x } { a } \right) + c \right]$
$ = \sqrt { 2 } \tan ^ { - 1 } \frac { \left( t - \frac { 1 } { t } \right) } { \sqrt { 2 } } + C \quad \left[ \text { put } y = t - \frac { 1 } { t } \right]$
$ = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { t ^ { 2 } - 1 } { \sqrt { 2 } t } \right) + C$
$ = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { \tan x - 1 } { \sqrt { 2 \tan x } } \right) + C$ $[Put \ t^2 = tan x]$
$I = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { \tan x - 1 } { \sqrt { 2 \tan x } } \right) + C$
View full question & answer→Question 1011 Mark
Find $ \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
AnswerAccording to the question , $ I = \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
By Using integration by parts for first integral, we get
$ = \int {\log } (\mathop {\mathop{\rm l}\nolimits} \limits_Iogx) \cdot \mathop 1\limits_{II} dx + \int {\frac{1}{{{{(\log x)}^2}}}} dx$
$I = log (log x) \int 1 d x - \int \left[ \frac { d } { d x } \log ( \log x )\Big]\Big[ \int 1 d x \right] d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x$ $= x log (log x) - \int \frac { 1 } { ( \log x ) } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x)-\int (\mathop {\mathop{\rm\frac { 1}{logx}}\nolimits} \limits_I ) \cdot \mathop 1\limits_{II} dx + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
By Using integration by parts for second integral, we get
$= x log (log x) -\frac{x}{logx}+ \int [\frac {d}{dx}(\frac {1}{logx})]. (\int 1dx)dx+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x \frac {1}{x} . x+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$ = x \log ( \log x )-\frac{x}{logx}+ C$
$\therefore I = x \log ( \log x ) -\frac{x}{logx}+ C$
View full question & answer→Question 1021 Mark
Find $\int \frac{x^{4} d x}{(x-1)\left(x^{2}+1\right)}$
AnswerWe have $\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}=(x+1)+\frac{1}{x^{3}-x^{2}+x-1}$
Using Partitial Fraction,
= $(x+1)+\frac{1}{(x-1)\left(x^{2}+1\right)}$ ......(i)
Now express $\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^{2}+1\right)}$ .....(ii)
So $1 = A (x^2 + 1) + (Bx + C) (x - 1)$
$= (A + B) x^2 + (C - B) x + A - C$
Equating coefficients on both sides, we get A + B = 0, C - B = 0 and A - C = 1, which gives
$A=\frac{1}{2}, B=C=-\frac{1}{2}$
Substituting values of A, B and C in (ii), we get
$\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}$ .....(iii)
Again, substituting (iii) in (i), we have
$\frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)}$ = $(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}$
Therefore
$\int \frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} d x$ = $\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+C$
View full question & answer→Question 1031 Mark
Find $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$
AnswerWe have $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\int \frac{\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}}{x^{4}} d x$
Put, $1-\frac{1}{x^{3}}=1-x^{-3}=t$ $\Rightarrow \frac{3}{x^{4}} d x=d t$
Therefore $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\frac{1}{3} \int t^{\frac{1}{4}} d t=\frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}}+C=\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{\frac{5}{4}}+C$
View full question & answer→Question 1041 Mark
Find $\int \cos 6 x \sqrt{1+\sin 6 x} d x$
AnswerPut t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore $\int \cos 6 x \sqrt{1+\sin 6 x} d x$ = $\frac{1}{6} \int t^{\frac{1}{2}} d t$
= $\frac{1}{6} \times \frac{2}{3}(t)^{\frac{3}{2}}+\mathrm{C}=\frac{1}{9}(1+\sin 6 x)^{\frac{3}{2}}+\mathrm{C}$
View full question & answer→Question 1051 Mark
Evaluate $\int_{0}^{\frac{\pi}{2}} \log \sin x d x$
AnswerLet $I=\int_{0}^{\frac{\pi}{2}} \log \sin x d x$
$I=\int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\frac{\pi}{2}} \log \cos x d x$
Adding the two values of I, we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x$
= $\int_{0}^{\frac{\pi}{2}}(\log \sin x \cos x+\log 2-\log 2) d x$ (by adding and subtracting log 2)
= $\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x$
Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when $x=\frac{\pi}{2}$, t = $\pi$
Therefore, 2I = $\frac{1}{2} \int_{0}^{\pi} \log \sin t d t-\frac{\pi}{2} \log 2$
= $\frac{2}{2} \int_{0}^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{2} \log 2$
= $\int_{0}^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{2} \log 2$ ((by changing variable t to x)
= $I-\frac{\pi}{2} \log 2$
Hence, $\int_{0}^{\frac{\pi}{2}} \log \sin x d x=\frac{-\pi}{2} \log 2$
View full question & answer→Question 1061 Mark
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
Answer$I=\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }}} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\frac{{\sin x}}{{\cos x}}} }} = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx} } $...(1)
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}dx} $
$\left[ {\because \int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx} } } \right]$
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} $ ...(2)
Adding (1) and (2), we get
$2I = \int_{\pi /6}^{\pi /3} {1dx} $
$= \left[ x \right]_{\pi /6}^{\pi /3}$
$= \frac{\pi }{3} - \frac{\pi }{6} = \frac{\pi }{6}$
$I = \frac{\pi }{{12}}$
View full question & answer→Question 1071 Mark
Find the integral: $\int \frac{1-\sin x}{\cos ^{2} x} d x$
AnswerWe have
$\int \frac{1-\sin x}{\cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x} d x-\int \frac{\sin x}{\cos ^{2} x} d x$
= $\int \sec ^{2} x d x-\int \tan x \sec x d x$
= tan x - sec x + C
View full question & answer→Question 1081 Mark
Evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$
AnswerLet $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$ ...(i)
Then, by using, $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$, we get
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$ = $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x$ ...(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}$
Hence, $I=\frac{\pi}{4}$
View full question & answer→Question 1091 Mark
Find the integral: $\int cosec x(cosec x+\cot x) d x$
AnswerWe have
$\int\left(cosec x(cosec x+\cot x) d x=\int cosec ^{2} x d x+\int cosec x \cot x d x\right.$
= -cot x - cosec x + C
View full question & answer→Question 1101 Mark
Evaluate $\int\limits_{ - 1}^1 {{{\sin }^5} \ x{{\cos }^4} \ xdx} $
AnswerLet $f(x) = \sin^5x.\cos^4x$
$f(-x) = \sin^5(-x).\cos^4(-x)$
$= -\sin^5x.\cos^4x$
$= -f(x)$
$\therefore$ f is odd function
$\therefore \int\limits_{ - 1}^1 {{{\sin }^5}x.{{\cos }^4}xdx = 0} $
View full question & answer→Question 1111 Mark
Find the integral: $\int(\sin x+\cos x) d x$
AnswerWe have
$\int(\sin x+\cos x) d x=\int \sin x d x+\int \cos x d x$
= -cos x + sin x + C
View full question & answer→Question 1121 Mark
Evaluate $\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$.
AnswerAccording to the question , $I =\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$ .......(i)
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin ( \pi - x ) } { 1 + \cos ^ { 2 } ( \pi - x ) } d x$ $[\because \int _0^a f(x)= \int_0^a f(a-x)]$
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin x } { 1 + \cos ^ { 2 } x } d x$.....(ii)
On Adding eqs.(i) and (ii),we get
$2 I = \int _ { 0 } ^ { \pi } \frac { \pi \sin x } { \left( 1 + \cos ^ { 2 } x \right) } d x$
Put , $cos x = t$
$-sin x dx = dt$
$\Rightarrow sinx dx = -dt$
Lower limit , when x = 0, then t = 1
Upper limit ,when $x= \pi$, then t = -1
$= - \pi \int _ { 1 } ^ { - 1 } \frac { d t } { 1 + t ^ { 2 } }$
$\Rightarrow 2 I = \pi \int _ { - 1 } ^ { 1 } \frac { d t } { \left( 1 + t ^ { 2 } \right) } $
$= \pi \left[ \tan ^ { - 1 } t \right] _ { - 1 } ^ { 1 }$
$= \pi \left[ \frac { \pi } { 4 } - \left( - \frac { \pi } { 4 } \right) \right] $
$= \frac { \pi ^ { 2 } } { 4 }$
$ \therefore I = \frac { \pi ^ { 2 } } { 4 }$
View full question & answer→Question 1131 Mark
Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$
AnswerLet $f(x) = \sin^2x$
$f(-x) = \sin^2(-x) = \sin^2x = f(x)$
$\therefore$ function is even
$\therefore \int_{ - \pi /4}^{\pi /4} {{{\sin }^2}xdx = 2\int_0^{\pi /4} {{{\sin }^2}xdx} }$
$= \int_\limits0^{π/4} {2\left( {\frac{{1 - \cos 2x}}{2}} \right)dx} $
$= \int_\limits0^{π/4} {\left( {{{1 - \cos 2x}}{}} \right)dx} $
$= \left[ {x - \frac{{\sin 2x}}{2}} \right]_0^{\pi /4}$
$ = \frac{\pi }{4} - \frac{1}{2}$
View full question & answer→Question 1141 Mark
Evaluate $\int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } dx$
AnswerLet $I = \int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x$
Put $\tan ^ { - 1 } x = t \quad \Rightarrow \frac { 1 } { 1 + x ^ { 2 } } d x = d t$
Lower limit when x = 0, then t = 0
Upper limit when x = 1, then t = $\pi / 4.$
$\therefore I = \int _ { 0 } ^ { \pi / 4 } t d t = \left[ \frac { t ^ { 2 } } { 2 } \right] _ { 0 } ^ { \pi / 4 } = \frac { 1 } { 2 } \left[ \left( \frac { \pi } { 4 } \right) ^ { 2 } - ( 0 ) ^ { 2 } \right] = \frac { \pi ^ { 2 } } { 32 }$
View full question & answer→Question 1151 Mark
Evaluate $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$
AnswerPut $t = x^5 + 1$, then $dt = 5x^4 dx$.
Therefore, $\int 5 x^{4} \sqrt{x^{5}+1} d x$ = $\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^{5}+1\right)^{\frac{3}{2}}$
Hence, $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$ = $\frac{2}{3}\left[\left(x^{5}+1\right)^{\frac{3}{2}}\right]_{-1}^{1}$
= $\frac{2}{3}\left[\left(1^{5}+1\right)^{\frac{3}{2}}-\left((-1)^{5}+1\right)^{\frac{3}{2}}\right]$
= $\frac{2}{3}\left[2^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{2}{3}(2 \sqrt{2})=\frac{4 \sqrt{2}}{3}$
View full question & answer→Question 1161 Mark
Evaluate $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$
AnswerLet $I=\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$.
Consider $\int \sin ^{3} 2 t \cos 2 t d t$
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = $\frac{1}{2}$du
So $\int \sin ^{3} 2 t \cos 2 t d t=\frac{1}{2} \int u^{3} d u$
= $\frac{1}{8}\left[u^{4}\right]=\frac{1}{8} \sin ^{4} 2 t=\mathrm{F}(t)$
Therefore, by the second fundamental theorem of integral calculus
$I=F\left(\frac{\pi}{4}\right)-F(0)=\frac{1}{8}\left[\sin ^{4} \frac{\pi}{2}-\sin ^{4} 0\right]=\frac{1}{8}$
View full question & answer→Question 1171 Mark
Evaluate $\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$
AnswerLet $I=\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$
Using partial fraction, we get $\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}$
So, $\int \frac{x d x}{(x+1)(x+2)}$ = - log |x + 1| + 2log |x + 2| = F(x)
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) - F(1) = [- log 3 + 2 log 4] - [- log 2 + 2 log 3]
= -3 log 3 + log 2 + 2 log 4 = log$ \left(\frac{32}{27}\right)$
View full question & answer→Question 1181 Mark
Evaluate $\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$
AnswerLet $I=\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$.
We first find the antiderivative of the integrand.
Put $30-x^{\frac{3}{2}}=t$. Then $-\frac{3}{2} \sqrt{x} d x=d t$ or $\sqrt{x} d x=-\frac{2}{3} d t$
Thus, $\int \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$ = $-\frac{2}{3} \int \frac{d t}{t^{2}}=\frac{2}{3}\left[\frac{1}{t}\right]=\frac{2}{3}\left[\frac{1}{\left(30-x^{\frac{3}{2}}\right)}\right]=\mathrm{F}(x)$
Therefore, by the second fundamental theorem of calculus, we have
I = F(9) - F(4)
= $\frac{2}{3}\left[\frac{1}{(30-27)}-\frac{1}{30-8}\right]$
$=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right]=\frac{19}{99}$
View full question & answer→Question 1191 Mark
Evaluate $\int_{2}^{3} x^{2} d x$
Answerf(x) is continuous in [2, 3]
$\int_{a}^{b} f(x) d x$ = $\mathop {\lim }\limits_{n \to \infty } h\sum\limits_{r = 0}^{n - 1} f (a + rh)$ where $\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$
Here,
$\int_{2}^{3}\left(x^{2}\right) d x$ = $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} f \left( {2 + \left( {\frac{r}{n}} \right)} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {{{\left( {2 + \left( {\frac{r}{n}} \right)} \right)}^2}}$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\frac{{{r^2}}}{{{n^2}}} + 4 + \frac{{4r}}{n}} \right)}$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{(n - 1)(n)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{4(n - 1)(n)}}{{2n}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {{n^2} - n} \right)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{2\left( {{n^2} - n} \right)}}{n}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left(\frac{\left(2 n^{3}-2 n^{2}-n^{2}+n\right)}{6 n^{2}}+4 n+\frac{2\left(n^{2}-n\right)}{n}\right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {2{n^3} - 3{n^2} + n} \right) + \left( {24{n^3}} \right) + \left( {12{n^3} - 12{n^2}} \right)}}{{6{n^2}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^2}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^3}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38}}{6}} \right) - \left( {\frac{{15}}{{6n}}} \right) + \left( {\frac{1}{{6{n^2}}}} \right)$
= $\frac{38}{6}$
= $\frac{19}{3}$
View full question & answer→Question 1201 Mark
Find $\int \sqrt{3-2 x-x^{2}} d x$
AnswerLet I = $\int \sqrt{3-2 x-x^{2}} \;d x$
= $\int \sqrt{-\left(x^{2}+2 x-3\right)} d x$
$=\int \sqrt{-\left(x^{2}+1+2 x-3-1\right)} d x$
= $\int \sqrt{-\left[(x+1)^{2}-4\right]} d x$
= $\int \sqrt{4-(x+1)^{2}} d x$
= $\int \sqrt{2^{2}-(x+1)^{2}} d x$ $\left[\because\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\sin^{-1}\left(\frac xa\right)+c\right]$
= $\frac{1}{2}(x+1) \sqrt{4-(x+1)^{2}} +\frac{4}{2} \sin ^{-1}\left(\frac{x+1}{2}\right)+c$
$\therefore \ \mathrm{I}=\int \sqrt{2^{2}-(x+1)^{2}} d x$
$=\frac{1}{2}(x+1) \sqrt{3-2 x-x^{2}}+2 \sin ^{-1} \frac{(x+1)}{2}+c$
View full question & answer→Question 1211 Mark
Find $\int \sqrt{x^{2}+2 x+5} d x$
AnswerNote that
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{(x+1)^{2}+4} d x$
Put x + 1 = y, so that dx = dy. Then
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{y^{2}+2^{2}} d y$
= $\frac{1}{2} y \sqrt{y^{2}+4}+\frac{4}{2} \log |y+\sqrt{y^{2}+4}|+C$
= $\frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log |x+1+\sqrt{x^{2}+2 x+5}|+\mathrm{C}$
View full question & answer→Question 1221 Mark
Find the integral: $\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x$
AnswerWe have
$\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x=\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x$
= $\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C$
= $\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C$
View full question & answer→Question 1231 Mark
Find $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$
AnswerWe have $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$
Consider $f(x) = \tan^{-1} x$, then f′(x) = $\frac{1}{1+x^{2}}$
Thus, the given integrand is of the form $e^x [f(x) + f′(x)]$.
Therefore, $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x = e^x \tan^{-1} x + C$
View full question & answer→Question 1241 Mark
Find the integral: $\int\left(x^{\frac{2}{3}}+1\right) d x$
AnswerWe have
$\int\left(x^{\frac{2}{3}}+1\right) d x=\int x^{\frac{2}{3}} d x+\int 1d x$
= $\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+\mathrm{C}=\frac{3}{5} x^{\frac{5}{3}}+x+\mathrm{C}$
View full question & answer→Question 1251 Mark
Find $\int e^{x} \sin x d x$
AnswerTake $e^x$ as the first function and sin x as second function.
Using Integrating by parts, we have
$\mathrm{I}=\int e^{x} \sin x d x=e^{x}(-\cos x)+\int e^{x} \cos x d x$
$= -e^x \cos x + I_1$ ......(i)
Taking $e^x$ and cos x as the first and second functions, respectively, in $I_1$, we get
$I_{1}=e^{x} \sin x-\int e^{x} \sin x d x$
Substituting the value of $I_1$ in (i), we get
$I = -e^x \cos x + e^x sin x - I$
$\Rightarrow~2I = e^x$ (sin x - cos x)
Hence, $I=\int e^{x} \sin x d x=\frac{e^{x}}{2}(\sin x-\cos x)+C$
View full question & answer→Question 1261 Mark
Find the integral: $\int \frac{x^{3}-1}{x^{2}} d x$
AnswerWe have
$\int \frac{x^{3}-1}{x^{2}} d x=\int x d x-\int x^{-2} d x$
= $\left(\frac{x^{1+1}}{1+1}+C_{1}\right)-\left(\frac{x^{-2+1}}{-2+1}+C_{2}\right); C_1, C_2$ are constants of integration
= $\frac{x^{2}}{2}+C_{1}-\frac{x^{-1}}{-1}-C_{2}$ = $\frac{x^{2}}{2}+\frac{1}{x}+C_{1}-C_{2}$
= $\frac{x^{2}}{2}+\frac{1}{x}+C$ where $C = C_1 - C_2$ is another constant of integration
View full question & answer→Question 1271 Mark
Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$
AnswerLet first function be $\sin^{-1} x$ and second function be $\frac{x}{\sqrt{1-x^{2}}}$.
First we find the integral of the second function, i.e., $\int \frac{x d x}{\sqrt{1-x^{2}}}$
Put $t =1 - x^2$. Then dt = -2x dx
Therefore, $\int \frac{x d x}{\sqrt{1-x^{2}}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}$
Hence, $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$ = $(\sin ^{-1} x)$$\int \frac{x d x}{\sqrt{1-x^{2}}}$ $-\int [\frac d{dx} sin^{-1}x.\int \frac{x d x}{\sqrt{1-x^{2}}} ] d x$
= $\left(\sin ^{-1} x\right)(-\sqrt{1-x^{2}})-\int \frac{1}{\sqrt{1-x^{2}}}(-\sqrt{1-x^{2}}) d x$
= $-\sqrt{1-x^{2}} \sin ^{-1} x+x+C$
= $x-\sqrt{1-x^{2}} \sin ^{-1} x+\mathrm{C}$
View full question & answer→Question 1281 Mark
Find $\int x e^{x} d x$
AnswerTake first function as x and second function as $e^x$. The integral of the second function is $e^x$
Therefore, $\int x e^{x} d x=x e^{x}-\int 1 \cdot e^{x} d x=x e^{x}-e^{x}+C$
View full question & answer→Question 1291 Mark
Find $\int \log x d x$
AnswerWe take log x as the first function and the constant function 1 as the second function.
Hence, $\int(\log x ).1 d x=\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x$
= $(\log x) \cdot x-\int \frac{1}{x} x d x=x \log x-x+C$
View full question & answer→Question 1301 Mark
Find $\int x \cos x d x$
AnswerPut f (x) = x (first function) and g (x) = cos x (second function).
Then, integration by parts gives
$\int x \cos x d x=x \int \cos x d x-\int\left[\frac{d}{d x}(x) \int \cos x d x\right] d x$
= $x \sin x-\int \sin x d x=x \sin x+\cos x+\mathrm{C}$
View full question & answer→Question 1311 Mark
Find: $\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
AnswerLet, $I=\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
By partial fractions
$\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}$
$x^2 + x + 1 = A(x^2+ 1) + Bx + C(x + 2)$
$=Ax^2+ A + Bx^2+ 2Bx + Cx + 2Cx^2+ x + 1$
$= x^2(A + B) + x(2B + C) + A + 2C$
On comparing coefficients of equation $(i),$ we get
$1 = A + B$ and $1 = 2B + C$
On solving above three equations, we get
$A=\frac{3}{5}$, $ B=\frac{2}{5} $ and $C=\frac{1}{5}$
Hence, $\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{x^{2}+1}$
Therefore, $\int \frac{\left(x^{2}+x+1\right) d x}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5} \int \frac{1}{x+2} d x+\frac{2}{5} \int \frac{x}{x^{2}+1} d x +\frac{1}{5} \int \frac{1}{\left(x^{2}+1\right)} d x$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \int \frac{2 x}{x^{2}+1}+\frac{1}{5} \int \frac{d x}{x^{2}+1}$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+c$
View full question & answer→Question 1321 Mark
Find: $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$
AnswerLet $y=\sin \phi$ $\Rightarrow d y=\cos \phi d \phi$
Therefore, $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$ = $\int \frac{(3 y-2) d y}{5-\left(1-y^{2}\right)-4 y}$
= $\int \frac{3 y-2}{y^{2}-4 y+4} d y$
Now, we write $\frac{3 y-2}{(y-2)^{2}}=\frac{A}{y-2}+\frac{B}{(y-2)^{2}}$
Therefore, 3y - 2 = A (y - 2) + B
Comparing the coefficients of y and constant term,
we get A = 3 and B = 4. for y = 2, y= 0
Therefore, the required integral is given by
$I=\int\left[\frac{3}{y-2}+\frac{4}{(y-2)^{2}}\right] d y$ = $3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^{2}}$
= $3 \log |y-2|+4\left(-\frac{1}{y-2}\right)+C$
= $3 \log |\sin \phi-2|+\frac{4}{2-\sin \phi}+C$
= $3 \log (2-\sin \phi)+\frac{4}{2-\sin \phi}+C$ (since, sin $\phi $ $\in$[-1,1], sin $\phi $< 2, 2 - sin $\phi $ is always positive)
View full question & answer→Question 1331 Mark
Find: $\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$
AnswerConsider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put $x^2 = y$
Then, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$
Write, $\frac{y}{(y+1)(y+4)}=\frac{A}{y+1}+\frac{B}{y+4}$
So that, y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides,
we get A + B = 1 and 4A + B = 0, which give
$A=-\frac{1}{3}$ and $B=\frac{4}{3}$
Thus, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3\left(x^{2}+1\right)}+\frac{4}{3\left(x^{2}+4\right)}$
= $-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$
= $-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+C$
View full question & answer→Question 1341 Mark
Find: $\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
AnswerWe write
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
So that $3x - 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)^2$
$= A (x^2 + 4x + 3) + B (x + 3) + C (x^2 + 2x + 1)$
Comparing coefficient of $x^2, x$ and constant term on both sides,
we get $A + C = 0, 4A + B + 2C = 3$ and $3A + 3B + C = -2$. Solving these equations, we get
$A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$. Thus the integrand is given by
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Therefore, $\int \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
= $\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C}$
= $\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$
View full question & answer→Question 1351 Mark
Write an anti derivative of function using the method of inspection: $\frac{1}{x}, x \neq 0$
AnswerWe know that
$\frac{d}{d x}(\log x)=\frac{1}{x}, x>0$ and $\frac{d}{d x}[\log (-x)]=\frac{1}{-x}(-1)=\frac{1}{x}, x<0$
Combining above, we get $\frac{d}{d x}(\log |x|)=\frac{1}{x}, x \neq 0$
Therefore, $\int \frac{1}{x} d x=\log |x|$ is one of the anti derivatives of $\frac{1}{x}$
View full question & answer→Question 1361 Mark
Find $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
AnswerHere the integrand $\frac{x^{2}+1}{x^{2}-5 x+6}$ is not proper rational function, so we divide $x^2 + 1$ by $x^2 - 5x + 6$ and find that
$\frac{x^{2}+1}{x^{2}-5 x+6}=1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}$
Let $\frac{5 x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
So that 5x - 5 = A (x - 3) + B (x - 2)
Equating the coefficients of x and constant terms on both sides,
we get A + B = 5 and 3A + 2B = 5.
Solving these equations, we get A = -5 and B = 10
Thus, $\frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$
Therefore, $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}$
= x - 5 log | x - 2| + 10 log | x - 3| + C.
View full question & answer→Question 1371 Mark
Write an anti derivative of function using the method of inspection: $3x^2 + 4x^3$
AnswerA function whose anti derivative is $3x^2 + 4x^3$.
$\frac{d}{d x}\left(x^{3}+x^{4}\right)=3 x^{2}+4 x^{3}$
Therefore, an anti derivative of $3x^2 + 4x^3$ is $x^3 + x^4$
View full question & answer→Question 1381 Mark
Find $\int \frac{d x}{(x+1)(x+2)}$
AnswerThe integral is a proper rational function.
Therefore, by using the form of partial fraction, we write
$\frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}$ ......(i)
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A + B = 0 and 2A + B = 1
Solving these equations, we get A = 1 and B = -1.
Thus, the integral is given by
$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}$
Therefore, $\int \frac{d x}{(x+1)(x+2)}=\int \frac{d x}{x+1}-\int \frac{d x}{x+2}$
= log |x + 1| - log |x + 2| + C
= $\log \left|\frac{x+1}{x+2}\right|+C$
View full question & answer→Question 1391 Mark
Write an anti derivative of function cos 2x using the method of inspection.
AnswerA function whose derivative is cos 2x.
$\frac{d}{d x} \sin 2 x=2 \cos 2 x$
or $\cos 2 x=\frac{1}{2} \frac{d}{d x}(\sin 2 x)=\frac{d}{d x}\left(\frac{1}{2} \sin 2 x\right)$
Therefore, an anti derivative of cos 2x is $\frac{1}{2} \sin 2 x$
View full question & answer→Question 1401 Mark
Find the integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
AnswerGiven integral is:$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
Let us express
$x+3=A \frac{d}{d x}\left(5-4 x-x^{2}\right)+B$ = A (- 4 - 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
- 2A = 1 and -4 A + B = 3, i.e., $A=-\frac{1}{2}$ and B = 1
Therefore, $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}+\int \frac{d x}{\sqrt{5-4 x-x^{2}}}$
= $-\frac{1}{2} I_{1}+I_{2}$ ........(i)
In $I_1$, put $5 - 4x - x^2 = t$, so that (- 4 - 2x) dx = dt.
Therefore, $I_{1}=\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+C_{1}$
= $2 \sqrt{5-4 x-x^{2}}+\mathrm{C}_{1}$ .......(ii)
Now consider $I_{2}=\int \frac{d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d x}{\sqrt{9-(x+2)^{2}}}$
Put x + 2 = t, so that dx = dt.
Therefore, $I_{2}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1} \frac{t}{3}+C_{2}$
= $\sin ^{-1} \frac{x+2}{3}+C_{2}$ .......(iii)
Substituting (ii) and (iii) in (i), we obtain
$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}}=-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, $ where $\mathrm{C}=\mathrm{C}_{2}-\frac{\mathrm{C}_{1}}{2}$
View full question & answer→Question 1411 Mark
Find the integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$
AnswerLet,$ I = ∫\frac{(x+2)}{(2x^2+6x+5)}dx$
we express
$x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B} = A(4x + 6) + B$
Equating the coefficients of x and the constant terms from both sides, we get
$4A = 1$ and $6A + B = 2$ or $A=\frac{1}{4}$ and $B=\frac{1}{2}$
Therefore, $\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}$
$= \frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2}$ .....(i)
In $I_1,$ put $2x^2 + 6x + 5 = t,$ so that $(4x + 6) dx = dt$
Therefore, $I_{1}=\int \frac{d t}{t}=\log |t|+C_{1}$
$= \log |2x^2 + 6x + 5| + C .....(ii)$
and $I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}}$
$= \frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$
Put $x+\frac{3}{2}=t$, so that $dx = dt,$ we get
$I_{2}=\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+C_{2}$
$\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2}......(iii)$
Using (ii) and (iii) in (i), we get
$\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\mathrm{C}$
where, $C=\frac{C_{1}}{4}+\frac{C_{2}}{2}$
View full question & answer→