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} $
$= \frac{{{x^3}}}{3}\log x - \frac{1}{3}\int{x^2}dx$
$= \frac{{{x^3}}}{3}\log x - \frac{1}{3}\frac{{{x^3}}}{3} + c$
$= \frac{{{x^3}}}{3}\log x - \frac{{{x^3}}}{9} + c$
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