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^{4 }x = \tan^{2 }x.\tan^{2 }x$
$= (\sec^{2 }x - 1) \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - (\sec^{2 }x - 1)$
$= \sec^{2 }x \tan^{2 }x - \sec^{2 }x + 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 $\tan x = 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^2x \sin^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 $(1 – \tan x) = t$
$\Rightarrow -\sec^2 x\ dx = dt$
$\therefore ~ \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x=\int \frac{-d t}{t^2}=-\int t^{-2} d t$
$= \frac{1}{t}+C$
$= \frac{1}{(1-\tan x)}+C$
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}}$
$\Rightarrow - 8x\ d\ x = dt$
$\therefore$ 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→