Question 13 Marks
Evaluate the integral $\int\limits_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}dx} $ using substitution.
AnswerLet $I = \int\limits_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}dx} $…(i)
Putting 2x = t
$ \Rightarrow 2 = \frac{{dt}}{{dx}}$
$ \Rightarrow 2dx = dt$
$ \Rightarrow dx = \frac{{dt}}{2}$
Limits of integration when $x = 1,t = 2{\kern 1pt} \times 1 = 2$ and when $x = 2,t = 2 \times 2 = 4$
$\therefore$ From eq. (i),
$I = \int\limits_2^4 {\left( {\frac{1}{{\frac{t}{2}}} - \frac{1}{{2{{\left( {\frac{t}{2}} \right)}^2}}}} \right){e^t}\frac{{dt}}{2}} $
$= \int\limits_2^4 {\left( {\frac{2}{t} - \frac{2}{{{t^2}}}} \right){e^t}\frac{{dt}}{2}} $
$= \int\limits_2^4 {\frac{1}{2}.2\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right){e^t}{{dt}}{}} $
$= \int\limits_2^4 {\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right){e^t}{{dt}}{}} $
$= \int\limits_2^4 {\left\{ {f\left( t \right) + f'\left( t \right)} \right\}{e^t}{{dt}}{}} $
$= \left\{ {{e^t}f\left( t \right)} \right\}_2^4$
$= \left( {\frac{{{e^t}}}{t}} \right)_2^4$
$= \frac{{{e^4}}}{4} - \frac{{{e^2}}}{2}$
$= \frac{{{e^4} - 2{e^2}}}{4}$
View full question & answer→Question 23 Marks
Evaluate the integral $\int_{0}^{2} x \sqrt{x+2}$ (Put $x + 2 =t^2$) using substitution.
AnswerGiven integral is: $\int_{0}^{2} x \sqrt{x+2} d x$
Let $x + 2 = t^2 \Rightarrow$ dx = 2t dt
And $x = t^2 - 2$
when, x = 0, t = $\sqrt{2}$ and when x = 2, t = 2
So, $\int_{0}^{2} x \sqrt{x+2} d x=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t \cdot t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t^{2} d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t$
= $2\left[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{\sqrt{2}}^2$
= $2\left[\frac{(2)^{5}}{5}-\frac{2(2)^{3}}{3}-\frac{(\sqrt{2})^{5}}{5}+\frac{2(\sqrt{2})^{3}}{3}\right]$
= $2\left[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}\right]$
= $2\left[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\right]$
= $2\left[\frac{16+8 \sqrt{2}}{15}\right]$
= $\left[\frac{16(2+\sqrt{2})}{15}\right]$
= $\frac{16 \sqrt{2}(\sqrt{2}+1)}{15}$ .Which is the required solution.
View full question & answer→Question 33 Marks
Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi~ d \phi$ using substitution.
AnswerGiven: $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi$
Let $\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi \mathrm{d} \phi=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi \mathrm{d} \phi$
= $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(\cos ^{2} \phi\right)^{2} \cos \phi d \phi$
$\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^{2} \phi\right)^{2} \cos \phi d \phi$
Also, let $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$
when $\phi=0, \mathrm{t}=0$ and when $\phi=\frac{\pi}{2}, \mathrm{t}=1$
So, $I=\int_{0}^{1} \sqrt{t}\left(1-t^{2}\right)^{2} d t$
= $\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t$
= $\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right) d t$
= $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}$
= $\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
= $\frac{154+42-132}{231}=\frac{64}{231}$
View full question & answer→Question 43 Marks
Evaluate the definite integral $\int\limits_2^3 {\frac{{xdx}}{{{x^2} + 1}}} $
Answer$\int\limits_2^3 {\frac{{xdx}}{{{x^2} + 1}}} = \frac{1}{2}\int\limits_2^3 {\frac{{2x}}{{{x^2} + 1}}dx} $ $= \frac{1}{2}\left( {\log \left| {{x^2} + 1} \right|} \right)_2^3$
$= \frac{1}{2}\left( {\log \left| {10} \right| - \log \left| 5 \right|} \right)$
$= \frac{1}{2}\left( {\log 10 - \log 5} \right)$
$ = \frac{1}{2}\log \frac{{10}}{5}$
$= \frac{1}{2}\log 2$
View full question & answer→Question 53 Marks
Integrate the function $\sqrt {1 + \frac{{{x^2}}}{9}} $
Answer$\int {\sqrt {1 + \frac{{{x^2}}}{9}} } dx$ $= \int {\sqrt {\frac{{9 + {x^2}}}{9}} } dx$
$ = \int {\frac{{\sqrt {{x^2} + {3^2}} }}{3}} dx$
$= \frac{1}{3}\int {\sqrt {{x^2} + {3^2}} } dx$
$= \frac{1}{3}\left[ {\frac{x}{2}\sqrt {{x^2} + {3^2}} + \frac{{{3^2}}}{2}\log \left| {x + \sqrt {{x^2} + {3^2}} } \right|} \right] + c$
$\left[ {\because \int {\sqrt {{x^2} + {a^2}} dx} } \right.$ $\left. { = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + c} \right]$
$= \frac{x}{6}\sqrt {{x^2} + 9} + \frac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + c$
View full question & answer→Question 63 Marks
Integrate the function $\sqrt{x^{2}+3 x}$
Answer$I=\int \sqrt{x^{2}+3 x} d x$
= $\int \sqrt{x^{2}+3 x+\frac{9}{4}-\frac{9}{4}} d x$
= $\int \sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} d x$
We know that
$\Rightarrow$ $\int \sqrt{x^{2}-a^{2} x} d x$ = $\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
Therefore,
$I=\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x}\right|+C$
$=\frac{(2 \mathrm{x}+3)}{4} \sqrt{\mathrm{x}^{2}+3 \mathrm{x}}-\frac{9}{8} \log \left|\left(\mathrm{x}+\frac{3}{2}\right)+\sqrt{\mathrm{x}^{2}+3 \mathrm{x}}\right|+\mathrm{C}$
View full question & answer→Question 73 Marks
Integrate the function $\sqrt{x^{2}+4 x-5}$
Answer$I=\sqrt{x^{2}+4 x-5} d x$
= $\int \sqrt{\left(x^{2}+4 x+4\right)-9} d x$
= $\int \sqrt{(x+2)^{2}-(3)^{2}} d x$
We know that,
$\Rightarrow \int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
$\Rightarrow \mathrm{I}=\frac{(\mathrm{x}+2)}{2} \sqrt{\mathrm{x}^{2}+4 \mathrm{x}-5}-\frac{9}{2} \log |(\mathrm{x}+2)+\sqrt{\mathrm{x}^{2}+4 \mathrm{x}-5}|+\mathrm{C}$
View full question & answer→Question 83 Marks
Integrate the function $\sqrt{1-4 x-x^{2}}$
Answer$I=\int \sqrt{1-4 x-x^{2}} d x$
= $\int \sqrt{1-\left(x^{2}+4 x+4-4\right)} d x$
= $\int \sqrt{1+4-(x+2)^{2}} d x$
= $\int \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} d x$
We know that,
$\Rightarrow \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$
$\Rightarrow I=\frac{(\mathrm{x}+2)}{2} \sqrt{1-4 \mathrm{x}-\mathrm{x}^{2}}+\frac{5}{2} \sin ^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{5}}\right)+\mathrm{C}$
View full question & answer→Question 93 Marks
Integrate the function $\sqrt{x^{2}+4 x+1}$
AnswerLet $I=\int \sqrt{x^{2}+4 x+1} d x$
$=\int \sqrt{\left(x^{2}+4 x+4\right)-3} d x$
$= \int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x$
We know that
$\int \sqrt{(x)^{2}-(a)^{2}} d x$ = $\frac{x}{2} \sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
Therefore,
$\Rightarrow \mathrm{I}=\frac{(\mathrm{x}+2)}{2} \sqrt{\mathrm{x}^{2}+4 \mathrm{x}+1}-\frac{3}{2} \log |(\mathrm{x}+2)+\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+1}|+\mathrm{C}$
View full question & answer→Question 103 Marks
Integrate the function $\sqrt{x^{2}+4 x+6}$
Answer$I=\int \sqrt{x^{2}+4 x+6} d x$
= $\int \sqrt{x^{2}+4 x+4+2} d x$
= $\int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x$
We know that,
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$
Therefore,
$I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\frac{2}{2} \log |(x+2)+\sqrt{x^{2}+4 x+6}|+C$
= $\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log |(x+2)+\sqrt{x^{2}+4 x+6}|+C$
View full question & answer→Question 113 Marks
Integrate the function $\sqrt {1 - 4{x^2}} $
Answer$\int {\sqrt {1 - 4{x^2}} dx} $ $ = \int {\sqrt {{1^2} - ({2x})^2} dx} $
$= \frac{{\left( {\frac{{2x}}{2}} \right)\sqrt {{1^2} - {{\left( {2x} \right)}^2}} + \frac{{{1^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{{2x}}{1}} \right)}}{{ 2 {}}} + c$
$\left[ {\because \int {\sqrt {{a^2} - {x^2}} dx} } \right.$ $\left. { = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right]$
$= \frac{1}{2}\left[ {x\sqrt {1 - 4{x^2}} + \frac{1}{2}{{\sin }^{ - 1}}2x} \right] + c$
$= \frac{x}{2}\sqrt {1 - 4{x^2}} + \frac{1}{4}{\sin ^{ - 1}}2x + c$
$= \frac{1}{4}{\sin ^{ - 1}}2x + \frac{x}{2}\sqrt {1 - 4{x^2}} + c$
View full question & answer→Question 123 Marks
Integrate the rational function $\frac{x}{(x-1)^{2}(x+2)}$
AnswerLet $\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)}$
$\Rightarrow x = A(x - 1)(x + 2) + B(x + 2) + C( x - 1)^2$ …(i)
$\Rightarrow x = (A+C)x^2+ (A+3B-2C)x - 2A+2B+C$
Substituting x = 1 in equation (i), we get,
$B=\frac{1}{3}$
Equating the coefficients of $x^2$ and constant term, we get,
A + C = 0
-2A + 2B + C = 0
$A=\frac{2}{9}$ and $C=\frac{-2}{9}$
Thus,
$\frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)}$
$\Rightarrow$$\int \frac{x}{(x-1)^{2}(x+2)} d x=\int\left\{\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)}\right\} d x$
= $\frac{2}{9} \int \frac{1}{(x-1)} d x+\frac{1}{3} \int \frac{1}{(x-1)^{2}} d x-\frac{2}{9} \int \frac{1}{(x+2)} d x$
= $\frac{2}{9} \log |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9} \log |x+2|+C$
= $\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+\frac{1}{3}\left(\frac{-1}{x-1}\right)+C$
View full question & answer→Question 133 Marks
Integrate the rational function $\frac{1-x^{2}}{x(1-2 x)}$
AnswerOn dividing $1 - x^2$ by x(1 - 2x), we get,$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)$.......(i)
Now, let $\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}$
(2 - x) = A(1 - 2x) + Bx …...(ii)
Now, substituting x = 0 and $\frac{1}{2}$ in equation (ii), we get,
A = 2 and B = 3
Thus, $\frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{(1-2 x)}$
Now, putting this value in equation (ii), we get,
$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{(1-2 x)}\right)$
$\Rightarrow$$\int \frac{1-x^{2}}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{(1-2 x)}\right)\right\} d x$
= $\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C$
= $\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C$
View full question & answer→Question 143 Marks
Integrate the rational function $\frac{2 x}{x^{2}+3 x+2}$
AnswerLet $\frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$
$\Rightarrow$ 2x = A(x + 2)+ B(x + 1) …(i)
Substituting x = -1 and -2 respectively in equation (i), we get,
A = -2, B = 4
Thus,
$\frac{2 x}{x^{2}+3 x+2}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$
$\Rightarrow$ $\int \frac{2 x}{x^{2}+3 x+2} d x=\int\left\{\frac{-2}{(x+1)}+\frac{4}{(x+2)}\right\} d x$
= 4 log|x + 2| - 2 log|x + 1| + C
View full question & answer→Question 153 Marks
Integrate the rational function: $\frac{x}{(x-1)(x-2)(x-3)}$
AnswerLet $\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$
⇒ x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2) ......(i)
Substituting x = 1, 2 and 3 respectively in equation (1), we get,
$A=\frac{1}{2}, B=-2$ and $C=\frac{3}{2}$
Thus,
$\frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}$
$\Rightarrow$$\int \frac{x}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right\} d x$
= $\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+C$
View full question & answer→Question 163 Marks
Integrate the rational function $\frac{3 x-1}{(x-1)(x-2)(x-3)}$
AnswerLet $\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$
$\Rightarrow$ 3x -1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2) ........(i)
Substituting x = 1, 2 and 3 respectively in equation (i), we get,
A = 1, B = -5 and C = 4
Thus,
$\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}$
$\Rightarrow~\int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right\} d x$
= log|x - 1| - 5 log|x - 2| + 4 log|x - 3| + C
View full question & answer→Question 173 Marks
Integrate the rational function $\frac{1}{x\left(x^{4}-1\right)}$
AnswerGiven the function, $\frac{1}{x\left(x^{4}+1\right)}$
Multiplying numerator and denominator by $x^3$, we get,
$\frac{1}{x\left(x^{4}+1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}+1\right)}$
Therefore, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}+1\right)} d x$
Now, let $x^4 = t$
$4x^3dx = dt$
Thus, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$
1 = A(t - 1) + Bt …(i)
Substituting t = 0 and 1 in (i), we get
A = -1 and B = 1
Therefore $\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$
$\int \frac{1}{x\left(x^{4}+1\right)} d x=\frac{1}{4} \int \left\{\frac{-1}{t}+\frac{1}{t-1}\right\} d t$
= $\frac{1}{4}[-\log |t|+\log |t-1|]+C$
= $\frac{1}{4} \log \left|\frac{t-1}{t}\right|+C$
= $\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C$
View full question & answer→Question 183 Marks
Integrate the rational function $\frac{1}{{{x^2} - 9}}$
Answer$\int {\frac{1}{{{x^2} - 9}}dx} $ $=\int {\frac{1}{{{x^2} - {3^2}}}dx}$
$= \frac{1}{{2 \times 3}}\log \left| {\frac{{x - 3}}{{x + 3}}} \right| + c$
$\left[ {\because \int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2a}}\log \left| {\frac{{x - a}}{{x + a}}} \right|} } \right]$
$= \frac{1}{6}\log \left| {\frac{{x - 3}}{{x + 3}}} \right| + c$.
Which is the required solution.
View full question & answer→Question 193 Marks
Integrate the rational function $\frac{1}{x\left(x^{n}+1\right)}$ [Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n = t$]
AnswerGiven function is, $\frac{1}{x\left(x^{n}+1\right)}$
Multiplying numerator and denominator by $x^{n-1}$, we get,
$\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}$
Let $x^n = t$
$nx^{n-1}dx = dt$
Therefore,
$\int \frac{1}{x\left(x^{n}+1\right)} d x=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t$
Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}$
1 = A(1 + t) + Bt ...(i)
Substituting t = 0, -1 in equation (i), we get,
A =1 and B = -1
Thus,
$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(1+t)}$
$\int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n} \int\left\{\frac{1}{t}-\frac{1}{(1+t)}\right\} d t$
= $\frac{1}{n}[\log |t|-\log |t+1|]+C$
= $\frac{1}{n}\left[\log \left|x^{n}\right|-\log \left|x^{n}+1\right|\right]+C$
= $\frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+C$
View full question & answer→Question 203 Marks
Integrate the rational function $\frac{2}{(1-x)\left(1+x^{2}\right)}$
AnswerLet $\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{(1-x)}+\frac{Bx + C}{\left(1+x^{2}\right)}$
$\Rightarrow 2 = A(1 + x^2) + (Bx + C)(1 - x)$
$\Rightarrow 2 = A + Ax^2 + Bx - Bx^2 + C - Cx$
On comparing the coefficients of $x^2, x$ and constant term, we get,
A - B = 0
B - C = 0
A + C = 2
On solving these equations, we get,
A = 1, B =1 and C = 1
Thus,
$\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{1}{(1-x)}+\frac{x+1}{\left(1+x^{2}\right)}$
$\Rightarrow$$\int \frac{2}{(1-x)\left(1+x^{2}\right)} d x=\int \frac{1}{(1-x)} d x+\int \frac{x}{\left(1+x^{2}\right)} d x+\int \frac{1}{\left(1+x^{2}\right)} d x$
= $-\int \frac{1}{(x-1)} d x+\frac{1}{2} \int \frac{2 x}{\left(1+x^{2}\right)} d x+\int \frac{1}{\left(1+x^{2}\right)} d x$
= $-\log |x-1|+\frac{1}{2} \log \left|1+x^{2}\right|+\tan ^{-1} x+C$
View full question & answer→Question 213 Marks
Integrate the rational function $\frac{x^{3}+x+1}{x^{2}-1}$
AnswerGiven function is $\frac{x^{3}+x+1}{x^{2}-1}$
Dividing $(x^3 + x + 1)$ by $x^2 - 1$, we get,
$\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}$
Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$
Now, $2x + 1 = A(x – 1) + B(x + 1)$ ...(i)
Substituting x = 1 and -1 in equation (i), we get,
$A=\frac{1}{2}$ and $B=\frac{3}{2}$
Thus, $\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$
$\Rightarrow~~\int \frac{x^{3}+x+1}{x^{2}-1}=\int x d x+\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{2} \int \frac{1}{(x-1)} d x$
= $\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C$
View full question & answer→Question 223 Marks
Integrate the rational function $\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$
Answer$\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$$= \frac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}$
$= \frac{A}{{x + 1}} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}}$ ...(i)
$ \Rightarrow 5x = A(x + 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$
$\Rightarrow 5x = A(x^2 4) + B(x^2 - x - 2) + C(x^2 + 3x + 2)$
$\Rightarrow 2x = Ax^2 - 4A + Bx^2 - Bx - 2B + Cx^2 + 3Cx + 2C$
Comparing coefficients of $x^2: A + B + C = 0$ .......(ii)
Comparing coefficients of $x: B + 3C= 5$ .......(iii)
Comparing constants: $4A 2B + 2C = 0$ .......(iv)
On solving eq. (i), (ii) and (iii), we get $A = \frac{5}{3},B = \frac{{ - 5}}{2},C = \frac{5}{6}$
Putting the values of A, B and C in eq. (i),
$\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$
$= \frac{{\frac{5}{3}}}{{x + 1}} + \frac{{\frac{{ - 5}}{2}}}{{\left( {x + 2} \right)}} + \frac{{\frac{5}{6}}}{{x - 2}}$
$\therefore \int \frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}dx$$= \frac{5}{3}\int {\frac{1}{{x + 1}}dx - \frac{5}{2}\int {\frac{1}{{\left( {x + 2} \right)}}dx + \frac{5}{6}} \int {\frac{1}{{x - 2}}} } dx$
$= \frac{5}{3}\log \left| {x + 1} \right| - \frac{5}{2}\log \left| {x + 2} \right| + \frac{5}{6}\log \left| {x - 2} \right| + c$
View full question & answer→Question 233 Marks
Integrate the function $\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}$
AnswerLet $I = \int {\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} $
$= \frac{1}{3}\int {\frac{{{3x^2}}}{{\sqrt {{{\left( {{x^3}} \right)}^2} + {a^6}} }}dx} $ ...(i)
Putting $x^3 = t$
$\Rightarrow 3{x^2} = \frac{{dt}}{{dx}}$
$ \Rightarrow 3{x^2}dx = dt$
$\therefore$ From eq. (i),
$I = \frac{1}{3}\int \frac{dt}{\sqrt {t^2+a^6}}$
$I = \frac{1}{3}\int \frac{dt}{\sqrt {t^2+(a^3)^2}}$
$= \frac{1}{3}\log \left| {t + \sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} } \right| + c$
$= \frac{1}{3}\log \left| {{x^3} + \sqrt {{x^6} + {a^6}} } \right| + c$
View full question & answer→Question 243 Marks
Integrate the function $\frac{x^{2}}{1-x^{6}}$.
AnswerLet $x^3 = t$
$\Rightarrow 3x^2 dx = dt$
$\Rightarrow \int \frac{x^{2}}{1-x^{6}} d x=\frac{1}{3} \int \frac{d t}{1-t^{2}}$
$\Rightarrow\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]+C$
$\Rightarrow \frac{1}{6}\left[\log \left|\frac{1+x^{3}}{1-x^{3}}\right|\right]+C$
View full question & answer→Question 253 Marks
Integrate the function $\frac{3 x}{1+2 x^{4}}$
AnswerLet $\sqrt{2} x^{2}=t$
$\Rightarrow 2 \sqrt{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$
$\Rightarrow \int \frac{3 x}{1+2 x^{4}} d x=\frac{3}{2 \sqrt{2}} \int \frac{d t}{1+t^{2}}$
$\Rightarrow \frac{3}{2 \sqrt{2}}\left[\tan ^{-1} t\right]+C$
$=\frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} \mathrm{x}^{2}+\mathrm{C}$
View full question & answer→Question 263 Marks
Integrate the function $\frac{1}{\sqrt{9-25 x^{2}}}$
AnswerLet 5x = t
$\Rightarrow$ 5dx = dt
$\Rightarrow \int \frac{1}{\sqrt{9-25 x^{2}}} d x= \frac15\int \frac{d t}{\sqrt{3^{2}-t^{2}}}$
$\Rightarrow \frac{1}{5} \sin ^{-1}\left(\frac{t}{3}\right)+C$
$\Rightarrow \frac{1}{5} \sin ^{-1}\left(\frac{5 x}{3}\right)+C$
View full question & answer→Question 273 Marks
Integrate the function $\frac{1}{\sqrt{(2-x)^{2}+1}}$
AnswerLet 2 - x = t
$\Rightarrow$ -dx = dt
$\Rightarrow \int \frac{1}{\sqrt{(2-x)^{2}+1}} d x=-\int \frac{d t}{\sqrt{t^{2}+1}}$
$=-[\log |\mathrm{t}+\sqrt{\mathrm{t}^{2}+1}|]+\mathrm{C}$
$\begin{equation} =\log |t+\sqrt{t^{2}+1}|^{-1}+C \end{equation}$
$\begin{equation} =\log \frac{1}{|t+\sqrt{t^{2}+1}|}+C \end{equation}$
$=\log \left|\frac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}\right|+C$
View full question & answer→Question 283 Marks
Integrate the function $\frac{1}{{\sqrt {1 + 4{x^2}} }}$
Answer$\int {\frac{1}{{\sqrt {1 + 4{x^2}} }}} dx$ $= \int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} + {{\left( 1 \right)}^2}} }}} dx$
$= \frac{{\log \left| {\left( {2x} \right) + \sqrt {{{\left( {2x} \right)}^2} + {1^2}} } \right|}}{2}+c$ ...[dividing by 2 as coefficient of x is 2]
$\left[ {\because \int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|} } \right]$
$= \frac{1}{2}\log \left| {2x + \sqrt {4{x^2} + 1} } \right| + c$
View full question & answer→Question 293 Marks
Integrate the function $\frac{1}{{\sqrt {7 - 6x - {x^2}} }}$
Answer$\int {\frac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} $ $ = \int {\frac{1}{{\sqrt { - {x^2} - 6x + 7} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} + 6x - 7} \right)} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} + 6x + 9 - 9 - 7} \right)} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left\{ {{{\left( {x + 3} \right)}^2} - 16} \right\}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( 16 \right)}} - {{\left( {x + 3} \right)}^2}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( {x + 3} \right)}^2}} }}dx} $
$= {\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + c$ ...$\left[ {\because \int {\frac{1}{{{a^2} - {x^2}}}dx = {{\sin }^{ - 1}}\frac{x}{a}} } \right]$
View full question & answer→Question 303 Marks
Integrate the function $\frac{1}{\sqrt{x^{2}+2 x+2}}$
AnswerClearly, $\int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x$
Let x + 1 = t
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t$
$=\log |t+\sqrt{t^{2}+1}|+C$
$=\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C$
$=\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C$
View full question & answer→Question 313 Marks
Find the integrals of the function sin x sin 2x sin 3x
Answer$\int \sin x \sin 2 x \sin 3 x d x$ = $\int \sin x \cdot \frac{1}{2}[\{\cos (2 x-3 x)-\cos (2 x+3 x)\}] d x$
$= \frac{1}{2} \int\{\sin x \cos (-x)-\sin x \cos 5 x\} d x$
$= \frac{1}{2} \int\{\sin x \cos x-\sin x \cos 5 x\} d x$
$= \frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$
$= \frac{1}{4}\left[\frac{-\cos 2 x}{2}\right]-\frac{1}{2} \int \frac{1}{2}\left\{ \sin (x+5 x)+\sin (x-5 x)\right\} d x$
$= \frac{-\cos 2 x}{8}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$
$= \frac{-\cos 2 x}{8}-\frac{1}{4}\left[\frac{-\cos 6 x}{6}+\frac{\cos 4 x}{4}\right]+C$
$= \frac{-\cos 2 x}{8}-\frac{1}{8}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}\right]+C$
$= \frac{1}{8}\left[\frac{\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x\right]+C$
View full question & answer→Question 323 Marks
Find the integral of the function $\frac{\cos 2 x}{(\cos x+\sin x)^{2}}$
AnswerClearly, $\frac{\cos 2 x}{(\cos x+\sin x)^{2}}=\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}=\frac{\cos 2 x}{1+\sin 2 x}$
Now, $\int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x=\int \frac{\cos 2 x}{1+\sin 2 x} d x$
Let 1 + sin2x = t
$\Rightarrow$ 2cos2x dx = dt
Thus, $\int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x=\frac{1}{2} \int \frac{1}{t} d t$
$= \frac{1}{2} \log |\mathrm{t}|+\mathrm{C}$
$= \frac{1}{2} \log |1+\sin 2 x|+C$
$= \frac{1}{2} \log \left|(\cos x+\sin x)^{2}\right|+C$
= log|sinx + cosx| + C
View full question & answer→Question 333 Marks
Find the integrals of the function sin 3x cos 4x
Answer$\int {\sin 3x\cos 4xdx = \frac{1}{2}\int {2\sin 3x\cos 4xdx} }$ $ = \frac{1}{2}\int {\left\{ {\sin \left( {4x + 3x} \right) - \sin \left( {4x - 3x} \right)} \right\}dx} $ [Using 2 sin B cos A = sin (A + B) - sin (A - B)]
$= \frac{1}{2}\int {\left( {\sin 7x - \sin x} \right)dx} $
$ = \frac{1}{2}\left[ {\int {\sin 7xdx - \int {\sin xdx} } } \right]$
$= \frac{1}{2}\left[ {\frac{{ - \cos 7x}}{7} - \left( { - \cos x} \right)} \right] + c$
$= \frac{{ - 1}}{{14}}\cos 7x + \frac{1}{2}\cos x + c$
View full question & answer→Question 343 Marks
Find the integrals of the function $\frac{1}{\sin x \cos ^{3} x}$
AnswerClearly, $\frac{1}{\sin x \cos ^{3} x}=\frac{\sin x}{\cos ^{3} x}+\frac{1}{\sin x \cos x}$
$= \tan x \sec ^{2} x+\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x \cos x}{\cos ^{2} x}}$
$= \tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}$
Now, $\int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x$
Let tan x = t
$\Rightarrow \sec^2 x dx = dt$
$\Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x=\int \operatorname{td} t+\int \frac{1}{t} d t$
$= \frac{t^{2}}{2}+\log |t|+C$
$= \frac{1}{2} \tan ^{2} x+\log |\tan x|+C$
View full question & answer→Question 353 Marks
Find the integral of the function $\tan^4 x$
Answer$\tan^4x = \tan^2x.\tan^2x$
$= (\sec^2x - 1) \tan^2x$
$= \sec^2x \tan^2x - \tan^2x$
$= \sec^2x \tan^2x - (\sec^2x - 1)$
$= \sec^2x \tan^2x - \sec^2x + 1$
Now, $\int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 d x$
$= \int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C$
Now, let tanx = t
$\Rightarrow \sec^2x dx = dt$
$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$
$\Rightarrow \int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$
View full question & answer→Question 363 Marks
Find the integral of the function ${\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}}$
Answer$\int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} dx$ $= \int {\frac{{\left( {2{{\cos }^2}x - 1} \right) - \left( {2{{\cos }^2}\alpha - 1} \right)}}{{\left( {\cos x - \cos \alpha } \right)}}} dx$
$= \int {\frac{{\left( {2{{\cos }^2}x} \right) - \left( {2{{\cos }^2}\alpha} \right)}}{{\left( {\cos x - \cos \alpha } \right)}}} dx$
$ = \int {\frac{{2\left( {\cos^2 x - \cos^2 \alpha } \right)}}{{\left( {\cos x - \cos \alpha } \right)}}} dx$
$ = \int {\frac{{2\left( {\cos x + \cos \alpha } \right)\left( {\cos x - \cos \alpha } \right)}}{{\left( {\cos x - \cos \alpha } \right)}}} dx$
$ = \int 2({cos x + cos \alpha })\ dx$
$= 2\left( {\sin x + x\cos \alpha } \right) + c$.
View full question & answer→Question 373 Marks
Find the integral of the function $\cos^4 2x$
Answer$\cos^42x = (\cos^22x)^2$
$= \left(\frac{1+\cos 4 x}{2}\right)^{2}$
$= \frac{1}{4}\left[1+\cos ^{2} 4 x-2 \cos 4 x\right]$
$= \frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$
$= \frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 8 x+2 \cos 4 x\right]$
Now, $\int \cos ^{4} 2 x d x=\int\left[\frac{3}{8}+\frac{1}{8} \cos 8 x+\frac{1}{2} \cos 4 x\right] d x$
$= \frac{3 x}{8}+\frac{1}{64} \sin 8 x+\frac{1}{8} \sin 4 x+C$
View full question & answer→Question 383 Marks
Find the integral of the function $\sin^4 x$
AnswerWe can write, $\sin^4x = \sin^2xsin^2x$
$= \left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$
$=\frac{1}{4}(1-\cos 2 x)^{2}$
$= \frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$
$= \frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$
$= \frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$
$= \frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$
Now, $\int \sin ^{4} x d x=\frac{1}{4} \int\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right] d x$
$= \frac{1}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{\sin 4 x}{4}\right)-\frac{2 \sin 2 x}{2}\right]+C$
$= \frac{1}{8}\left[3 \mathrm{x}+\left(\frac{\sin 4 \mathrm{x}}{4}\right)-2 \sin 2 \mathrm{x}\right]+\mathrm{C}$
$= \frac{3 \mathrm{x}}{8}-\frac{1}{4} \sin 2 \mathrm{x}+\frac{1}{32} \sin 4 \mathrm{x}+\mathrm{C}$
View full question & answer→Question 393 Marks
Find the integrals of the function $\sin^2(2x + 5)$
Answer$\int {{{\sin }^2}\left( {2x + 5} \right)dx} $$= \int {\frac{1}{2}\left\{ {1 - \cos 2\left( {2x + 5} \right)} \right\}dx} $
Using ${\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}$
$ = \frac{1}{2}\int {\left\{ {1 - \cos \left( {4x + 10} \right)} \right\}dx} $
$= \frac{1}{2}\left[ {\int {1dx - \int {\cos \left( {4x + 10} \right)dx} } } \right]$
Using $\int {\cos \left( {ax + b} \right)dx = \frac{{\sin \left( {ax + b} \right)}}{a} + c} $
$= \frac{1}{2}\left[ {x - \frac{{\sin \left( {4x + 10} \right)}}{4}} \right] + c$
$= \frac{1}{2}x - \frac{1}{8}\sin \left( {4x + 10} \right) + c$
View full question & answer→Question 403 Marks
Integrate the function: $\sin (a x+b) \cos (a x+b)$
AnswerLet I = $\int \sin (a x+b) \cos (a x+b) d x$
We know that,
sin 2A = 2sinA.cosA
Therefore, sin (ax + b) cos (ax + b) = $\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$
Let 2(ax + b) = t
$\Rightarrow$ 2adx = dt
$\Rightarrow$ I = $\int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t}{2 a} d t$
= $\frac{1}{4 a}[-\cos t]+C$
=$\frac{-1}{4 a} \cos 2(a x+b)+C$
View full question & answer→Question 413 Marks
Integrate the function: $\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}$
AnswerLet $I = \int {\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}dx} $ ...(i) Putting x + log x = t
$ \Rightarrow 1 + \frac{1}{x} = \frac{{dt}}{{dx}}$
$ \Rightarrow \frac{{x + 1}}{x} = \frac{{dt}}{{dx}}$
$\Rightarrow \left( {\frac{{x + 1}}{x}} \right)dx = dt$
$\therefore$ From eq. (i), $I = \int {{t^2}dx} $
$= \frac{{{t^3}}}{3} + c$
$= \frac{1}{3}{\left( {x + \log x} \right)^3} + c$
View full question & answer→Question 423 Marks
Integrate the function: $\frac{1}{\cos ^{2} x(1-\tan x)^{2}}$
AnswerWe have $\frac{1}{\cos ^{2} x(1-\tan x)^{2}}=\frac{\sec ^{2} x}{(1-\tan x)^{2}}$
Let (1 – tanx) = t
$\Rightarrow -\sec^2xdx = dt$
$\therefore ~ \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x=\int \frac{-d t}{t^2}=-\int t^{-2} d t$
$= \frac{1}{t}+C$
$= \frac{1}{(1-\tan x)}+C$
View full question & answer→Question 433 Marks
Integrate the function: $\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}$
AnswerLet $I = \int {\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}dx} $ $ = \int {\frac{{2\cos x - 3\sin x}}{{2\left( {2\sin x + 3\cos x} \right)}}dx} $
$= \frac{1}{2}\int {\frac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx} $…(i)
Putting 2 sin x + 3 cos x = t
$ \Rightarrow 2\cos x - 3\sin x = \frac{{dt}}{{dx}}$
$ \Rightarrow $ (2 cos x - 3 sin x)dx = dt
$\therefore$ From eq. (i), $I = \frac{1}{2}\int {\frac{{dt}}{t} = \frac{1}{2}\log \left| t \right| + c} $
$= \frac{1}{2}\log \left| {2\sin x + 3\cos x} \right| + c$
View full question & answer→Question 443 Marks
Integrate the function: $\tan^2(2x - 3)$
Answer$\int {{{\tan }^2}\left( {2x - 3} \right)} dx$$ = \int {\left\{ {{{\sec }^2}\left( {2x - 3} \right) - 1} \right\}} dx$
$ = \int {{{\sec }^2}\left( {2x - 3} \right)} dx - \int {1dx} $
Using $\int {{{\sec }^2}\left( {ax + b} \right)} dx = \frac{{\tan \left( {ax + b} \right)}}{a} + c$
$= \frac{{\tan \left( {2x - 3} \right)}}{2} - x + c$
View full question & answer→Question 453 Marks
Integrated the function: $\int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x.$
AnswerLet $I = \int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x$
Put $e^{2x} + e^{-2x} = t$
$\Rightarrow \left( 2 e ^ { 2 x } - 2 e ^ { - 2 x } \right) d x = d t$ $\left[ \because \frac { d } { d x } \left( e ^ { a x } \right) = a e ^ { a x } \right]$
$\Rightarrow 2\left( e ^ { 2 x } - e ^ { - 2 x } \right) d x = { d t } $
$\Rightarrow \left( e ^ { 2 x } - e ^ { - 2 x } \right) d x = \frac { d t } { 2 }$
$\therefore I = \int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x$
$ = \frac { 1 } { 2 } \int \frac { d t } { t } $
$= \frac { 1 } { 2 } \log | t | + C$
$= \frac { 1 } { 2 } \log \left| e ^ { 2 x } + e ^ { - 2 x } \right| + C \quad \text { [ put } t = e ^ { 2 x } + e ^ { - 2 x } ]$
$\therefore I =\frac { 1 } { 2 } \log \left| e ^ { 2 x } + e ^ { - 2 x } \right| + C$
View full question & answer→Question 463 Marks
Integrate the function: $\frac{e^{2 x}~-1}{e^{2 x}~+1}$
AnswerWe have,
$\frac{e^{2 x}-1}{e^{2 x}+1}$
Dividing numerator and denominator by $e^x$, we get,
$\frac{\frac{e^{2 x}-1}{e^{x}}}{\frac{e^{2 x}+1}{e^{-x}}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
Let $e^x + e^{-x} = t$
Differentiating both sides, we get,
$(e^x - e^{-x} )dx = dt$
Now the integral becomes,
$= \int \frac{d t}{t}$
$= log |t| + C$
$= \log|e^x + e^{-x}| + C$
View full question & answer→Question 473 Marks
Integrate the function: $\frac{x}{e^{x^{2}}}$
AnswerLet $x^2 = t$
$\Rightarrow 2xdx = dt$
$\Rightarrow \int \frac{x}{e^{x^{2}}} d x=\frac{1}{2} \int \frac{1}{e^{t}} d t$
$=\frac{1}{2} \int e^{-t} d t$
$= \frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C$
$= \frac{-1}{2} e^{-x^{2}}+C$
$= \frac{-1}{2 e^{x^{2}}}+C$
View full question & answer→Question 483 Marks
Integrate the function: $ \frac{x}{{9 - 4{x^2}}}$
AnswerLet $I = \int {\frac{x}{{9 - 4{x^2}}}} dx$$ = \frac{{ - 1}}{8}\int {\frac{{ - 8x}}{{9 - 4{x^2}}}dx} $ ...(i)
Putting $9 - 4x^2 = t$
$ \Rightarrow - 8x = \frac{{dt}}{{dx}}$
$\Rightarrow - 8xdx = dt$
$\therefore$ From eq. (i), $I = \frac{{ - 1}}{8}\int {\frac{{dt}}{t} }$
$ = \frac{{ - 1}}{8}\log \left| t \right| + c$
$= \frac{{ - 1}}{8}\log \left| {9 - 4{x^2}} \right| + c$
View full question & answer→Question 493 Marks
Integrate the function: $\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}$
AnswerLet $I = \int {\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}dx} $$= \frac{1}{9}\int {\frac{{9{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}dx} $…(i)
Putting $2 + 3x^3 = t$
$ \Rightarrow 9{x^2} = \frac{{dt}}{{dx}}$
$\Rightarrow 9{x^2}dx = dt$
$\therefore $ From eq. (i), $I = \frac{1}{9}\int\frac{1}{t^3}dt$
$ = \frac{1}{9}\int {{t^{ - 3}}dt} $
$= \frac{1}{9}.\frac{{{t^{ - 2}}}}{{ - 2}} + c$
$= \frac{{ - 1}}{{18{t^2}}} + c$
$= \frac{{ - 1}}{{18{{\left( {2 + 3{x^3}} \right)}^2}}} + c$
View full question & answer→Question 503 Marks
Integrate the function: $\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}$
AnswerLet $x^3 - 1 = t$
$\Rightarrow 3x^2dx = dt$
$\Rightarrow\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{5} d x=\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} d x$
$= \int t^{\frac{1}{3}}(t+1) \frac{d t}{3}$
$=\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t$
$=\frac{1}{3}\left[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]+C$
$= \frac{1}{3}\left[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}\right]+C$
$= \frac{1}{7}\left(x^{3}-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^{3}-1\right)^{\frac{4}{3}}+C$
View full question & answer→Question 513 Marks
Integrate the function: $\frac{x}{\sqrt{x+4}}, x>0$
AnswerLet x + 4 = t
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{x}{\sqrt{x+4}} d x=\int \frac{(t-4)}{\sqrt{t}} d t$
= $\int\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right) d t$
= $\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$
= $\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C$
= $ \frac{2}{3} t \cdot t^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C$
= $\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C$
= $ \frac{2}{3} \sqrt{x+4}(x-8)+C$.
View full question & answer→Question 523 Marks
Find an anti derivative (or integral) of the function by the method of inspection sin $2x - 4e^{3x}$
AnswerWe know that, $\frac{d}{{dx}}\left( {\cos 2x} \right) = - 2\sin 2x$
$ \Rightarrow \frac{1}{{ - 2}}\frac{d}{{dx}}\left( {\cos 2x} \right) = \sin 2x$
$\Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x} \right) = \sin 2x$ ...(i)
Again $\frac{d}{{dx}}{e^{3x}} = 3{e^{3x}}$
$\Rightarrow \frac{d}{{dx}}\left( {\frac{1}{3}{e^{3x}}} \right) = {e^{3x}}$
$\frac{d}{{dx}}\left( {\frac{{ - 4}}{3}{e^{3x}}} \right) = - 4{e^{3x}}$ [Multiplying both sides by -4] ...(ii)
Adding eq. (i) and (ii), we get
$ \Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x} \right) + \frac{d}{{dx}}\left( {\frac{{ - 4}}{3}{e^{3x}}} \right) = sin\ 2x + (-4e^{3x})$
$ \Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = sin\ 2x - 4e^{3x}$
$\therefore $ An anti-derivative of sin $2x - 4e^{3x}$ is $\frac{{ - 1}}{2}\cos 2x - \frac{4}{3}{e^{3x}}.$
View full question & answer→Question 533 Marks
Find the integral: $\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$
Answer$\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$
= $\int\left(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}\right) d x$
= $3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
= $\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+C$
View full question & answer→Question 543 Marks
Find the integral: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
AnswerLet I =$\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
Separating the terms we get,
I =$\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x$
Applying the formula,
$\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
I = $\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+C$
= $\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+C$
= $\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C$
View full question & answer→Question 553 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^2 {x\sqrt {2 - x} dx} $
AnswerLet $I = \int\limits_0^2 {x\sqrt {2 - x} dx} $ $= \int\limits_0^2 {\left( {2 - x} \right)\sqrt {2 - \left( {2 - x} \right)} dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx } } } \right]$
$\Rightarrow I = \int\limits_0^2 {\left( {2 - x} \right)\sqrt x dx} $
$= \int\limits_0^2 {\left( {2{x^{\frac{1}{2}}} - {x^{\frac{3}{2}}}} \right)dx} $
$= \left[ {2.\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}}} \right]_0^2$
$= \left( {\frac{4}{3}{{.2}^{\frac{3}{2}}} - \frac{2}{5}{{.2}^{\frac{5}{2}}}} \right) - \left( {0 - 0} \right)$
$\Rightarrow I = \frac{4}{3} \times 2\sqrt 2 - \frac{2}{5} \times 4\sqrt 2 $
$ = \left( {\frac{8}{3} - \frac{8}{5}} \right)\sqrt 2 $
$= \frac{{16\sqrt 2 }}{{15}}$
View full question & answer→Question 563 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $
AnswerLet $I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $
$= \int\limits_0^1 {\left( {1 - x} \right)\left\{ {1 - {{\left( {1 - x} \right)}}} \right\}^ndx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx } } } \right]$
$\Rightarrow I = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $
$ = \int\limits_0^1 {\left( {1 - x} \right){x^n}dx} $
$= \int\limits_0^1 {\left( {{x^n} - {x^{n + 1}}} \right)dx} $
$\Rightarrow I = \left( {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{n + 2}}}}{{n + 2}}} \right)_0^1$
$ = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} - \left( {0 - 0} \right)$
$= \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$
View full question & answer→Question 573 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_2^8 {\left| {x - 5} \right|dx} $
AnswerLet $I = \int\limits_2^8 {\left| {x - 5} \right|dx} $…(i)
Putting x - 5 = 0
$ \Rightarrow x = 5 \in \left( {2,8} \right)$
$\therefore$ From eq. (i),
$I = \int\limits_2^5 {\left| {x - 5} \right|dx + \int\limits_5^8 {\left| {x + 5} \right|dx} } $
$= \int\limits_2^5 { - \left( {x - 5} \right)dx + \int\limits_5^8 {\left( {x - 5} \right)dx} } $
$= - \left( {\frac{{{x^2}}}{2} - 5x} \right)_2^5 + \left( {\frac{{{x^2}}}{2} - 5x} \right)_{^5}^8$
$= - \left[ {\left( {\frac{{25}}{2} - 25} \right) - \left( {2 - 10} \right)} \right] + \left[ {\left( {32 - 40} \right) - \left( {\frac{{25}}{2} - 25} \right)} \right]$
$= - \left( { - \frac{{25}}{2} + 8} \right) + \left( { - 8 + \frac{{25}}{2}} \right)$
$= \frac{{25}}{2} - 8 - 8 + \frac{{25}}{2}$
= 25 - 16
= 9
View full question & answer→Question 583 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_{ - 5}^5 {\left| {x + 2} \right|dx} $
AnswerLet $I = \int\limits_{ - 5}^5 {\left| {x + 2} \right|dx} $ …(i) Putting x + 2 = 0
$ \Rightarrow x = - 2 \in \left( { - 5,5} \right)$
$\therefore$ From eq. (i),
$I = \int\limits_{ - 5}^{ - 2} {\left| {x + 2} \right|} dx + \int\limits_{ - 2}^5 {\left| {x + 2} \right|dx} $
$= \int\limits_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + \int\limits_{ - 2}^5 {\left( {x + 2} \right)dx} } $
$= - \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 5}^{ - 2} + \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 2}^5$
$= - \left[ {\left( {\frac{4}{2} - 4} \right) - \left( {\frac{{25}}{2} - 10} \right)} \right] $ $+ \left[ {\left( {\frac{{25}}{2} + 10} \right) - \left( {\frac{4}{2} - 4} \right)} \right]$
$= - \left( { - 2 - \frac{5}{2}} \right) + \left( {\frac{{45}}{2} + 2} \right)$
$= 2 + \frac{5}{2} + \frac{{45}}{2} + 2$
$ = 4 + 25 = 29$
View full question & answer→Question 593 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}} $
AnswerLet $I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^2}x + {{\cos }^5}x}}} dx$…(i)
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^5}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}dx} $
$\left[ {\because \int\limits_0^{{a }{}} {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)} dx } } \right]$
$\Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^5x}}}{{{{\cos }^5}x + {{\sin }^5}x}}dx} $ …(ii)
Adding equations (i) and (ii),
$2I = \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{{{\cos }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}} + \frac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}} \right)dx} $
$= \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{{{\cos }^5}x + {{\sin }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)dx} $
$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {1dx} $
$\Rightarrow 2I = \frac{\pi }{2}$
$\Rightarrow I = \frac{\pi }{4}$
View full question & answer→Question 603 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$
AnswerLet $I = \int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$…(i)
$\Rightarrow I = \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt {a - \left( {a - x} \right)} }}dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$= \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} dx$ …(ii)
Adding eq. (i) and (ii), $2I = \int\limits_0^a {\left( {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }} + \frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} \right)dx} $
$= \int\limits_0^a {\left( {\frac{{\sqrt x + \sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} \right)dx}$
$= \int\limits_0^a {1dx} = \left( x \right)_0^a = a$
$ \Rightarrow I = \frac{a}{2}$
View full question & answer→Question 613 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} dx$
AnswerLet $I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} $ …(i)
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}} dx$
$= \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
$= - \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \cos x\sin x}}} dx$…(ii)
Adding eq. (i) and (ii), we have $2I = 0 \Rightarrow I = 0$
View full question & answer→Question 623 Marks
By using the properties of definite integral, evaluate the integral $\int_{0}^{\pi} \frac{x d x}{1+\sin x}$
AnswerGiven, $\int_{0}^{\pi} \frac{\mathrm{x}}{1+\sin \mathrm{x}} \mathrm{dx}$
Let, $I=\int_{0}^{\pi} \frac{x}{1+\sin x} d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin x} d x$ .....(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})+\mathrm{x}}{1+\sin \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\int_{0}^{\pi} \frac{\pi}{1+\sin \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin \mathrm{x})}{(1+\sin \mathrm{x})(1-\sin \mathrm{x})} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin \mathrm{x})}{\cos ^{2} \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\frac{1}{\cos ^{2} \mathrm{x}}-\frac{\sin \mathrm{x}}{\cos ^{2} \mathrm{x}}\right\} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\sec ^{2} x-\tan x \sec x\right\} d x$
$\Rightarrow 2 \mathrm{I}=\pi[\tan \mathrm{x}-\sec \mathrm{x}]_{0}^{\pi}$
$\Rightarrow 2 \mathrm{I}=\pi[2]$
$\Rightarrow I=\pi$
View full question & answer→Question 633 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $
AnswerLet $I = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $…(i)
$= \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}\left( {\frac{\pi }{2} - x} \right)dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$ \Rightarrow I= \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} $ …(ii)
Adding equation (i) and (ii),
$2I = \int\limits_0^{\frac{\pi }{2}} {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)dx} $
$= \int\limits_0^{\frac{\pi }{2}} {1dx} $
$\Rightarrow 2I = \left( x \right)_0^{\frac{\pi }{2}}$
$ \Rightarrow 2I = \frac{\pi }{2}$
$ \Rightarrow I = \frac{\pi }{4}$
View full question & answer→