Question 511 Mark
Integrate the function x log 2x
Answer$\int {x\log 2xdx} $ $ = \int {\left( {\log 2x} \right)xdx} $
$= \left( {\log 2x} \right)\int {xdx - \int {\left[ {\frac{d}{{dx}}\log 2x\int {xdx} } \right]dx} } $
[Applying product rule]
$= \left( {\log 2x} \right)\frac{{{x^2}}}{2} - \int {\frac{1}{{2x}}.2.\frac{{{x^2}}}{2}dx} $
$ = \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\int {xdx} $
$= \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\frac{{{x^2}}}{2} + c$
$= \frac{{{x^2}}}{2}\log 2x - \frac{{{x^2}}}{4} + c$
View full question & answer→Question 521 Mark
Integrate the function x log x
AnswerLet $ I = x log x$
Now, integrating by parts, we get,
Taking, Logarithmic function as first function and algebraic function as second function,
$I=\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x$
= $\log x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$
= $\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x$
= $\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C$
View full question & answer→Question 531 Mark
Integrate the function $x^2e^x$
Answer$\int {{x^2}{e^x}} dx$$ = {x^2}\int {{e^x}dx - \int {\left[ {\frac{d}{{dx}}{x^2}\int {{e^x}dx} } \right]} dx} $
[Applying product rule]
$ = {x^2}{e^x} - \int {2x{e^x}dx} $
$ = {x^2}{e^x} - 2\int {x{e^x}dx} $
$= {x^2}{e^x} - 2\left[ {{xe^x}dx - \int {\left\{ {\frac{d}{{dx}}x\int {{e^x}dx} } \right\}dx} } \right]$
[Again applying product rule]
$= {x^2}{e^x} - 2\left( {x{e^x} - \int {1.{e^x}dx} } \right)$
$= {x^2}{e^x} - 2\left( {x{e^x} - \int {{e^x}dx} } \right)$
$ = {x^2}{e^x} - 2x{e^x} + 2\int {{e^x}dx} $
$= x^2e^x - 2xe^x + 2e^x + c$
$= e^x(x^2 - 2x + 2) + c$
View full question & answer→Question 541 Mark
$\int e^{x} \sec x(1+\tan x) d x$ equals
Answer$\int e^{x} \sec x(1+\tan x) d x$
Let
$\mathrm{I}=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x ...(i)$
Also, let $sec x = f(x)$
$\Rightarrow \sec x \tan x = f’(x)$
We know that $\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+C$
Thus $(i)$ gives, $I = e^{x }sec x + C$
View full question & answer→Question 551 Mark
$\int x^{2} e^{x^{3}} d x$ equals
AnswerLet $I=\int x^{2} e^{x^{3}} d x$
Also, let $x^3 = t, $
$\Rightarrow 3x^2dx = dt$
Thus,
$\Rightarrow I=\frac{1}{3} \int e^{t} d t$
$=\frac{1}{3}\left(e^{t}\right)+C$
$=\frac{1}{3}\left(e^{x^{3}}\right)+C$
View full question & answer→Question 561 Mark
Integrate the function ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$
AnswerPutting $x = \tan \theta $
$ \Rightarrow dx = {\sec ^2}\theta d\theta $
$\therefore \int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} $
$= \int {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right).{{\sec }^2}\theta d\theta }$
$= \int {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right).{{\sec }^2}\theta d\theta } $
$= \int {2\theta {{\sec }^2}\theta d\theta }$
$ = 2\int {\theta {{\sec }^2}\theta d\theta } $ ...[Applying product rule]
$= 2\left[ {\theta .\tan \theta - \int {1.\tan \theta d\theta } } \right]$
$= 2\left[ {\theta .\tan \theta - \int {\tan \theta d\theta } } \right]$
$ = 2\left[ {\theta \tan \theta - \log \sec \theta } \right] + c$
$= 2\left[ {{{\tan }^{ - 1}}x.x - \log \sqrt {1 + {x^2}} } \right] + c$
$= 2\left[ {x{{\tan }^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right)} \right] + c$
$= 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + c$
View full question & answer→Question 571 Mark
Integrate the function $e^{2x} \sin x$
AnswerLet $ = e^{2x }\sin xI$
Integrating by parts, we get,
$I=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d\ x\right\} d\ x$
$= \sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d\ x$
$= \frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x\ d x$
Again, integrating by parts, we get,
$I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \int e^{2 x} d\ x-\int\left\{\left(\frac{d}{d x} \cos x\right) \int e^{2 x} d x\right\} d x\right]$
$= \frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \cdot \frac{e^{2 x}}{2} d x\right]$
$=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I$
$\Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow \frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]$
View full question & answer→Question 581 Mark
Integrate the function $\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}$
AnswerLet $I = \frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx$
$= \int {\frac{{\left( {x - 1} \right) - 2}}{{{{\left( {x - 1} \right)}^3}}}{e^x}} dx$
$= \int {{e^x}\left[ {\frac{{\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^3}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\Rightarrow I = \int {{e^x}\left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} } \right]$
It is in the form of ${\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} }$ since here $f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}$ and $f'\left( x \right) = \frac{d}{{dx}}\left\{ {{{\left( {x - 1} \right)}^{ - 2}}} \right\}$
$ = - 2{\left( {x - 1} \right)^{ - 3}}$
$ = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}$.
$ \Rightarrow I = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$
View full question & answer→Question 591 Mark
Integrate the function x sin 3x
Answer$\int {x\sin 3xdx} $ $= x\int {\sin 3xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin 3x} dx} \right)dx} } $
[Applying product rule]
$= x\left( {\frac{{ - \cos 3x}}{3}} \right) - \int {1\left( {\frac{{ - \cos 3x}}{3}} \right)dx + c}$
$ = \frac{{ - 1}}{3}x\cos 3x + \frac{1}{3}\int {\cos 3xdx + c} $
$= \frac{{ - 1}}{3}x\cos 3x + \frac{1}{3}\frac{{\sin 3x}}{3} + c$
$= \frac{{ - 1}}{3}x\cos 3x + \frac{1}{9}\sin 3x + c$
$= - \frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + c$
View full question & answer→Question 601 Mark
Integrate the function ${e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$
AnswerLet $I = {e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx$ $\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} } \right]$
It is in the form of ${\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} }$, Here $f\left( x \right) = \frac{1}{x} = {x^{ - 1}}$ and $f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}$
$ \Rightarrow I = {e^x}\frac{1}{x} + c$
$= \frac{{{e^x}}}{x} + c$ ...$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx = {e^x}f\left( x \right) + c} \right]$
View full question & answer→Question 611 Mark
Integrate the function ${e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right)$
AnswerLet $I = \int {{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}dx} $ $ = \int {{e^x}.\frac{{1 + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}dx} $
$ = \int {{e^x}\left[ {\frac{1}{{2{{\cos }^2}\frac{x}{2}}} + \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right]dx} $
$= \int {{e^x}\left( {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right)dx}$
$= \int {{e^x}\left( {\tan \frac{x}{2} + \frac{1}{2}{{\sec }^2}\frac{x}{2}} \right)} dx$
$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx} \right]$
It is in the form of $\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx$ since here $f\left( x \right) = \tan \frac{x}{2}$ and $f'\left( x \right) = \frac{1}{2}{\sec ^2}\frac{x}{2}$
$ = {e^x}\tan \frac{x}{2} + c$
$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$
View full question & answer→Question 621 Mark
Integrate the function $\frac{x e^{x}}{(1+x)^{2}}$
Answer$I=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{\frac{x}{(1+x)^{2}}\right\} d x$
= $\int e^{x}\left\{\frac{1+x-1}{(1+x)^{2}}\right\} d x$
= $\int e^{x}\left\{\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right\} d x$
Now,
Let $\frac{1}{1+x}=f(x) \Rightarrow f^{\prime}(x)=-\frac{1}{(1+x)^{2}}$
We know that,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C$
Thus,
$\int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+C$
View full question & answer→Question 631 Mark
Integrate the function $e^x (\sin x + \cos x)$
Answer$I=\int e^{x}(\sin x+\cos x) d x$
Now,
Let $\sin x = f(x) $
$\Rightarrow f'(x) = \cos x $
We know that,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Thus,
$\int e^{x}(\sin x+\cos x) d x=e^{x} \sin x+C$
View full question & answer→Question 641 Mark
Integrate the function $(x^2 + 1) \log x$
Answer$\int {\left( {{x^2} + 1} \right)\log xdx} $
$ = \int {\left( {\log x} \right)\left( {{x^2} + 1} \right)dx} $
[Applying product rule]
$= \log x\left( {\frac{{{x^3}}}{3} + x} \right) - \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \int {\left( {\frac{{{x^2}}}{3} + 1} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {{x^2}dx - \int {1dx} } $
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\frac{{{x^3}}}{3} - x + c$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{{{x^3}}}{9} - x + c$
View full question & answer→Question 651 Mark
Integrate the function $x (\log x)^2$
AnswerLet $I=x(\log x)^{2}$
Integrating by parts, we get,
$I=\left[(\log x)^2 \int x d x-\int\left\{\left(\frac{d}{d x} (\log x)^2 \right) \int x d x\right\} d x\right]$
$= \left[\frac{x^{2}}{2}(\log x)^2-\int \frac{2 \log x }{x} \cdot \frac{x^{2}}{2} d x\right]$
$= \frac{x^{2}}{2}(\log x)^{2}- \int x \log x \cdot d x$
$= \frac{x^{2}}{2}(\log x)^{2} - [ \log x \int x dx -\int( \frac {d}{dx} (\log x)\int x dx) dx]$
$= \frac{x^{2}}{2}(\log x)^{2} - [ \frac {x²}{2}\log x - \int(\frac1x \cdot\frac{x^2}{2} ) dx$
$= \frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C$
View full question & answer→Question 661 Mark
Integrate the function $\tan^{-1}x$
AnswerLet $I = \tan^{-1}xdx$
$= \int {\left( {{{\tan }^{ - 1}}x} \right).1} dx$
$ = {\tan ^{ - 1}}x.x - \int {\frac{1}{{1 + {x^2}}}x.dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {\left( {1 + {x^2}} \right)} \right| + c$
$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \log \left| {f\left( x \right)} \right|} } \right]$
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + c$
View full question & answer→Question 671 Mark
Integrate the function $x \sec^2 x$
AnswerLet I $= x \sec^{2 }x$
Now, integrating by parts, we get,
$I=x \int \sec ^{2} x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sec ^{2} x d x\right\} d x$
$= x \tan x-\int (1) (\tan x) d x$
$= x \tan x+\log |\cos x|+c$
View full question & answer→Question 681 Mark
Integrate the function: $\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$
AnswerLet $I=\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$
$I=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$
Now, integrating by parts, we get,
$I = \frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right]$
= $-\frac{1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right]$
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x\right]$
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+C$
= $-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+C$
View full question & answer→Question 691 Mark
Integrate the function $(\sin^{-1}x)^2$
AnswerPutting $x = \sin \theta $
$\Rightarrow dx = \cos \theta d\theta $
$\therefore \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx$
$= \int {{\theta ^2}\cos \theta } d\theta $
$[$Applying product rule$]$
$ = {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta } $
$= {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta }$
$[$Again applying product rule$]$
$= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]$
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta d\theta } $
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\sin \theta + c$
$= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c$
View full question & answer→Question 701 Mark
Integrate the function x sin x
Answer$\int {x\sin x} dx$ $= x\int {\sin x} dx - \int {\left( {\frac{d}{{dx}}x\int {\sin xdx} } \right)dx} $
[Applying product rule]
$ = x\left( { - \cos x} \right) - \int {1\left( { - \cos x} \right)dx} $
$ = -x\cos x - \int { - \cos xdx} $
$= - x\cos x + \int {\cos xdx} $
= -x cos x + sin x + c
View full question & answer→Question 711 Mark
$\int \frac{d x}{x\left(x^{2}+1\right)}$ equals
AnswerLet $\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$
$1 = A(X^2 + 1) + (Bx + C)x = Ax^2 + A + Bx^2 + Cx = (A + B)x^2+Cx + A$
Equating the coefficients of $x^2, x$ and constant term, we get,
$A + B = 0$
$C = 0$
$A = 1$
On solving these equations, we get,
$A = 1, B = -1$ and $C = 0$
Therefore, $\frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
$\int \frac{1}{x\left(x^{2}+1\right)}=\int\left\{\frac{1}{x}+\frac{-x}{x^{2}+1}\right\} d x$
$= \log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C$
View full question & answer→Question 721 Mark
$\int \frac{x d x}{(x-1)(x-2)}$ equals
AnswerLet $\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$
x = A(x - 2) + B(x - 1) …(i)
Substituting x = 1 and 2 in (i), we get,
A = -1 and B = 2
Therefore, $\frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$
$\int \frac{x}{(x-1)(x-2)} d x=\int\left\{-\frac{1}{(x-1)}+\frac{2}{(x-2)}\right\} d x$
= $-\log |x-1|+2 \log |x-2|+c$
= $\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
View full question & answer→Question 731 Mark
$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$ equals
Answer$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}=\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x\right)}}$
$=\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64}\right)}}$
$=\int \frac{d x}{\sqrt{-4\left[\left(x-\frac{9}{8}\right)^{2}-\left(\frac{9}{8}\right)^{2}\right.}]}$
$=\frac{1}{2}\left[\sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)\right]+C$
$=\frac{1}{2}\left[\sin ^{-1}\left(\frac{8 x-9}{9}\right)\right]+C$
View full question & answer→Question 741 Mark
$\int \frac{d x}{x^{2}+2 x+2}$ equals
Answer$\int \frac{d x}{x^{2}+2 x+2}=\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}$
$= \int \frac{1}{(x+1)^{2}+(1)^{2}} d x$
$=\left[\tan ^{-1}(x+1)\right]+C$
View full question & answer→Question 751 Mark
$\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals
AnswerLet I = $\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$
$=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x~~~$
$=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$
$= \int \sec ^{2} x d x+\int cosec ^{2} x d x$
$$= tanx - cotx + C
View full question & answer→Question 761 Mark
$\int \frac{10 x^{9}+10^{x} \log _{e} 10 d x}{x^{10}+10^{x}}$, equals
AnswerLet $x^{10} + 10^x = t$
$\Rightarrow (10x^9 + 10^x \log_e10)dx = dt$
$\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$
$= \log t + C$
$= \log(x^{10} + 10^x) + C$
View full question & answer→Question 771 Mark
If $\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$ such that f(2) = 0. Then f(x) is
AnswerIt if given that $\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$
$f(x)=\int 4 x^{3}-\frac{3}{x^{4}} d x$
$\Rightarrow f(x)=4 \int x^{3} d x-3 \int\left(x^{-4}\right) d x$
$\Rightarrow f(x)=4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C$
$\Rightarrow f(x)=x^{4}+\frac{1}{x^{3}}+C$
Also, It is given that f(2) = 0
$\Rightarrow f(2)=(2)^{4}+\frac{1}{(2)^{3}}+\mathrm{C}=0$
$\Rightarrow 16+\frac{1}{8}+C=0$
$\Rightarrow \mathrm{C}=-\left(16+\frac{1}{8}\right)=-\frac{129}{8}$
Therefore, $f(x)=x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$. Which is the required solution.
View full question & answer→Question 781 Mark
The anti derivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
Answer$\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
= $\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x$
= $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$
= $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C$
View full question & answer→Question 791 Mark
The value of $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is
AnswerGiven Integral is: $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ ......(i)
As, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-\mathrm{x}\right)}{4+3 \cos \left(\frac{\pi}{2}-\mathrm{x}\right)}\right) \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos \mathrm{x}}{4+3 \sin x}\right) \mathrm{d} \mathrm{x}$ ....(ii)
Adding (i) and (ii), we get
$~~~~~2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left\{\log \left(\frac{4+3 \sin \mathrm{x}}{4+3 \cos \mathrm{x}}\right)+\left(\frac{4+3 \cos \mathrm{x}}{4+3 \sin }\right)\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \mathrm{x}}{4+3 \cos \mathrm{x}} \times \frac{4+3 \cos \mathrm{x}}{4+3 \sin x }\right) \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log 1 \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} 0 . \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=0$
View full question & answer→Question 801 Mark
The value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$ is
AnswerGiven: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$
Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$
$\Rightarrow \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\mathrm{x}^{3}\right) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{x} \cos \mathrm{x}) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\tan ^{5} \mathrm{x}\right) \mathrm{dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1) \mathrm{dx}$
It is also known that if f(x) is an even function then,
$\Rightarrow \mathrm{I}=0+0+0+2 \cdot \int_{0}^{\frac{\pi}{2}}(1) \mathrm{d} \mathrm{x}$ ...$\left\{\int_{-a}^{a} f(x) d x=0\right\}$
$\Rightarrow \mathrm{I}=2 \cdot[\mathrm{x}]_{0}^{\frac{\pi}{2}}$
$\Rightarrow I=2 \cdot \frac{\pi}{2}$
$\Rightarrow \mathrm{I}=\pi$
View full question & answer→Question 811 Mark
Find the integral: $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}$
AnswerWe have $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}$
= $\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$ (completing the square)
Put $x-\frac{1}{5}=t$. Then dx = dt
Therefore, $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}$
= $\frac{1}{\sqrt{5}} \log |t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}|+C$
= $\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+C$
View full question & answer→Question 821 Mark
Find the integral: $\int \frac{d x}{3 x^{2}+13 x-10}$
AnswerWe write the denominator of the integrand,
$3x^2 + 13x - 10 = 3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)$
$= 3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] ($completing the square$)$
Thus, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$
Put $x+\frac{13}{6}=t$. Then $dx = dt$
Therefore, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}}$
$= \frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+\mathrm{C}_{1}$
$= \frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1}$
$= \frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1}$
$= \frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3}$
$= \frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C$ where, $\mathrm{C}=\mathrm{C}_{1}+\frac{1}{17} \log \frac{1}{3}$
View full question & answer→Question 831 Mark
Find the integral: $\int \frac{d x}{x^{2}-6 x+13}$
AnswerWe have $x^2 - 6x + 13 $
$= x^2 - 6x + 3^2 - 3^2 + 13 $
$= (x - 3)^2 + 4$
So, $\int \frac{d x}{x^{2}-6 x+13}$
$=\int \frac{1}{(x-3)^{2}+2^{2}} d x$
Let $x – 3 = t$
$\Rightarrow dx = dt$
Therefore, $\int \frac{d x}{x^{2}-6 x+13}$
$=\int \frac{d t}{t^{2}+2^{2}}$
$=\frac{1}{2} \tan ^{-1} \frac{t}{2}+\mathrm{C}$
$=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+C$
View full question & answer→Question 841 Mark
Find the integral: $\int \frac{d x}{\sqrt{2 x-x^{2}}}$
AnswerLet I = $\int \frac{d x}{\sqrt{2 x-x^{2}}}$
$= \int \frac{d x}{\sqrt{1-(x-1)^{2}}}$
Put $x - 1 = t.$ Then $dx = dt.$
Therefore, $\int \frac{d x}{\sqrt{2 x-x^{2}}}=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1}(t)+C$
$= \sin^{-1} (x-1) + C$
View full question & answer→Question 851 Mark
Find the integral: $\int \frac{d x}{x^{2}-16}$
AnswerWe have $\int \frac{d x}{x^{2}-16}=\int \frac{d x}{x^{2}-4^{2}}=\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|+C$$~~~(Using \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C)$
View full question & answer→Question 861 Mark
Find: $\int \sin ^{3} x d x$
AnswerFrom the identity $\sin 3x = 3 \sin x - 4 \sin^3 x,$ we find that
$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$
Therefore, $\int \sin ^{3} x d x=\frac{3}{4} \int \sin x d x-\frac{1}{4} \int \sin 3 x d x$
Integral = $-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x+c$
View full question & answer→Question 871 Mark
Find: $\int \sin 2 x \cos 3 x d x$
AnswerAs we know , sin x cos y = $\frac{1}{2}[\sin (x+y)+\sin (x-y)]$
Then , $\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]$
= $\frac{1}{2}\left[-\frac{1}{5} \cos 5 x+\cos x\right]+C$
= $-\frac{1}{10} \cos 5 x+\frac{1}{2} \cos x+C$
View full question & answer→Question 881 Mark
Find: $\int \cos ^{2} x d x$
Answer$\cos 2x = 2 \cos^2 x - 1$
$\cos ^{2} x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^{2} x d x=\frac{1}{2} \int(1+\cos 2 x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$
$=\frac{x}{2}+\frac{1}{4} \sin 2 x+C$
View full question & answer→Question 891 Mark
Find the integral: $\int \frac{1}{1+\tan x} d x$
Answer$\int \frac{d x}{1+\tan x}=\int \frac{\cos x d x}{\cos x+\sin x}$
= $\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x}$
= $\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
= $\frac{x}{2}+\frac{C_{1}}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$ ...(i)
Now, consider I = $\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
Put cos x + sin x = t so that (cos x - sin x) dx = dt
Therefore, $\mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{2}=\log |\cos x+\sin x|+\mathrm{C}_{2}$
Putting it in (i), we get
$\int \frac{d x}{1+\tan x}=\frac{x}{2}+\frac{\mathrm{C}_{1}}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{2}}{2}$
= $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{C_{1}}{2}+\frac{C_{2}}{2}$
= $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C},\left(\mathrm{C}=\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2}\right)$
View full question & answer→Question 901 Mark
Find the integral: $\int \frac{\sin x}{\sin (x+a)} d x$
AnswerPut $x + a = t.$ Then $dx = dt.$ Therefore
$\int \frac{\sin x}{\sin (x+a)} d x=\int \frac{\sin (t-a)}{\sin t} d t$
$= \int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t$
$= \cos a \int d t-\sin a \int \cot t d t$
$= (\cos a) t-(\sin a)\left[\log |\sin t|+C_{1}\right]$
$= (\cos a)(x+a)-(\sin a)\left[\log |\sin (x+a)|+C_{1}\right]$
$= x \cos a + a \cos a - (\sin a) \log |\sin (x + a)| - C_1 \sin a$
Hence, $\int \frac{\sin x}{\sin (x+a)} d x = x \cos a - (\sin a) \log |\sin (x + a)| - C, $
where, $C = -C_1 \sin a + a \cos a,$ is another arbitrary constant.
View full question & answer→Question 911 Mark
Find the integral: $\int \sin ^{3} x \cos ^{2} x d x$
AnswerWe have
$\int \sin ^{3} x \cos ^{2} x d x=\int \sin ^{2} x \cos ^{2} x(\sin x) d x$
= $\int\left(1-\cos ^{2} x\right) \cos ^{2} x(\sin x) d x$
Put t = cos x so that dt = -sin x dx
Therefore, $\int \sin ^{2} x \cos ^{2} x(\sin x) d x$ = $-\int\left(1-t^{2}\right) t^{2} d t$
= $-\int\left(t^{2}-t^{4}\right) d t=-\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)+\mathrm{C}$
= $-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+C$
View full question & answer→Question 921 Mark
Integrate the function $w.r.t. x: \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}$
AnswerDerivative of $\tan ^{-1} x=\frac{1}{1+x^{2}}$.
Thus, we use the substitution $\tan^{-1} x = t$
so that $\frac{d x}{1+x^{2}}=d\ t$
Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d\ x$
$=\int \sin t\ d\ t = -\cos t + C$
$ = -\cos(\tan^{-1}x) + C$
View full question & answer→Question 931 Mark
Integrate the function w.r.t. x: $\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}$
AnswerDerivative of $\sqrt{x}$ is $\frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}}$.
Thus, we use the substitution $\sqrt{x}=t$ so that $\frac{1}{2 \sqrt{x}} d x=d t$ giving $dx = 2t\ dt.$
Thus, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x$ = $\int \frac{2 t \tan ^{4} t \sec ^{2} t d t}{t}$ = $2 \int \tan ^{4} t \sec ^{2} t d t$
Again, we make another substitution $\tan t = u$ so that $\sec^2 t dt = du$
Therefore, $2 \int \tan ^{4} t \sec ^{2} t\ d t=2 \int u^{4} d u=2 \frac{u^{5}}{5}+C$
$= \frac{2}{5} \tan ^{5} t+C ($since $u = \tan t)$
$= \frac{2}{5} \tan ^{5} \sqrt{x}+C$ (since $t=\sqrt{x}$)
Hence, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x=\frac{2}{5} \tan ^{5} \sqrt{x}+C$
View full question & answer→Question 941 Mark
Integrate the function w.r.t. x: $2x \sin (x^2 + 1)$
AnswerDerivative of $x^2 + 1$ is $2x.$
Thus, we use the substitution $x^2 + 1 = t$
so that $2x\ dx = dt.$
Therefore, $\int 2 x \sin \left(x^{2}+1\right) d x=\int \sin t d t=-\cos t+C=-\cos \left(x^{2}+1\right)+C$
View full question & answer→Question 951 Mark
Integrate the function sin mx w.r.t. x.
AnswerWe know that derivative of mx is m.
Thus, we make the substitution mx = t so that mdx = dt.
Therefore, $\int \sin m x d x=\frac{1}{m} \int \sin t d t$ $=-\frac{1}{m} \cos t+\mathrm{C}=-\frac{1}{m} \cos m x+\mathrm{C}$
View full question & answer→Question 961 Mark
Evaluate $ \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x.$
AnswerAccording to the question , $ I = \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(i)
$ \Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } ( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x ) } d x$$ \left[ \because \int _ { 0 } ^ { a } f ( x ) d x = \int _ { 0 } ^ { a } f ( a - x ) d x \right]$
$ \Rightarrow \quad I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(ii)
On adding Equations (i) and (ii) we get ,
$ 2 I = \int _ { 0 } ^ { \pi } \frac { ( x + \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$
$ \Rightarrow \quad 2 I = \pi \int _ { 0 } ^ { \pi } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
we know that,$ \int _ { 0 } ^ { 2 a } f ( x ) d x = 2 \int _ { 0 } ^ { a }f(x) dx \ , \ if \ \ f(2a - x) = f(x)$
Here, $ a^2 cos^2( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x )$
$= a^2 cos^2 x + b^2 sin^2x$
$ \therefore \quad 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
On dividing numerator and denominator by $cos^2x$ we get ,
$ 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { \sec ^ { 2 } x } { a ^ { 2 } + b ^ { 2 } \tan ^ { 2 } x } d x$
Put $tan x = t \implies sec^2 x dx = dt$
Lower limit when x = 0, then $t = tan 0 = 0$
Upper limit when x = $ \frac { \pi } { 2 },$then $ t =\tan \frac { \pi } { 2 } = \infty$
$ \therefore \quad I = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + b ^ { 2 } t ^ { 2 } }$
$ = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + ( b t ) ^ { 2 } }$ $ = \frac { \pi } { b ^ { 2 } } \int _ { 0 } ^ { \infty } \frac { d t } { \left( \frac { a } { b } \right) ^ { 2 } + t ^ { 2 } }$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \frac { b t } { a } \right] _ { 0 } ^ { \infty } \left[ \because \int \frac { d x } { a ^ { 2 } + x ^ { 2 } } = \frac { 1 } { a } \tan ^ { - 1 } \frac { x } { a } + C \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \infty - \tan ^ { - 1 } 0 \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \frac { \pi } { 2 } - 0 \right]$$ \left[ \begin{array} { c } { \because \tan ^ { - 1 } \infty = \tan ^ { - 1 } \left( \tan \frac { \pi } { 2 } \right) = \frac { \pi } { 2 } } \\ { \text { and } \tan ^ { - 1 } 0 = \tan ^ { - 1 } \left( \tan 0 ^ { \circ } \right) = 0 } \end{array} \right]$
$ \therefore I = \frac { \pi ^ { 2 } } { 2 a b }$
View full question & answer→Question 971 Mark
Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin \ (\pi x)| d x$
AnswerHere $f (x) = | x \sin \pi x| = \left\{\begin{array}{l} {x \sin \pi x \text { for }-1 \leq x \leq 1} \\ {-x \sin \pi x \text { for } 1 \leq x \leq \frac{3}{2}} \end{array}\right.$
Therefore $\int_{-1}^{\frac{3}{2}}|x \sin \pi x| \ d x = \int_{-1}^{1} x \sin \pi x\ d x+\int_{1}^{\frac{3}{2}}-x \sin \pi x\ d x$
$= \int_{-1}^{1} x \sin \pi x\ d x-\int_{1}^{\frac{3}{2}} x \sin \pi x\ d x$
Integrating both integrals on righthand side, we get
$\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x = \left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{-1}^{1}-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{1}^{\frac{3}{2}}$
$= \frac{2}{\pi}-\left[-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right]$
$=\frac{3}{\pi}+\frac{1}{\pi^{2}}$
View full question & answer→Question 981 Mark
Find $\int \frac{\sin 2 x \cos 2 x d x}{\sqrt{9-\cos ^{4}(2 x)}}$
AnswerLet $I=\int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^{4} 2 x}} d x$
Put $\cos^2 (2x) = t$
so that $2 \sin 2x \cos 2x\ dx = -dt$
Therefore,
$I = -\frac{1}{2} \int \frac{d t}{\sqrt{9-t^{2}}}$
$= -\frac{1}{2} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C}=-\frac{1}{2} \sin ^{-1}\left[\frac{1}{3} \cos ^{2} 2 x\right]+\mathrm{C}$
View full question & answer→Question 991 Mark
Find the anti derivative $F$ of $f$ defined by $f(x) = 4x^3 - 6,$ where $F(0) = 3$
AnswerHere, $f(x) = x^4 - 6x$
Anti derivative$ F$ of $f(x)$ is given by;
$F(x)=\int f(x) d\ x$
$=\int\left(4 x^{3}-6\right) d\ x$
$=x^{4}-6 x+C$
So, $F(x)=x^{4}-6 x+C, $
where $C$ is constant of Integration.
Given: $F(0) = 3$, which gives,
$3 = 0 - 6 \times 0 + C$
$\Rightarrow C = 3$
Hence, the required anti derivative is the unique function $F$ defined by $F(x) = x^4 - 6x + 3$
View full question & answer→Question 1001 Mark
Find $ \int ( \sqrt { \cot x } + \sqrt { \tan x } )$dx
AnswerLet $I = \int [ \sqrt { \cot x } + \sqrt { \tan x } ] d x$ $=\int \frac{1}{\sqrt{tanx}} + \sqrt{tanx} \ [\because cotx = \frac {1}{tanx}]$
$ = \int \sqrt { \tan x } \Bigg[1 +\frac {1}{\Big(\sqrt { \tan x }\Big)^2}\Bigg] d x$
put $ tan x = t^2 \implies sec^2 x dx = 2t dt$
$ \Rightarrow \quad d x = \frac { 2 t } { sec^2x}dt$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 +tan^2x }dt$$[ \because 1 + \tan ^ { 2 } x = \sec ^ { 2 } x]$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 + (t ^ { 2 })^2 }$$[ tan x = t^2 ]$
$ \Rightarrow \quad d x = \frac { 2 t } { 1 + t ^ { 4 } }$
$ \therefore \quad I = \int t \left( 1 + \frac { 1 } { t ^ { 2 } } \right) \frac { 2 t } { \left( 1 + t ^ { 4 } \right) } d t [\because tanx =t^2 \implies \sqrt {tanx} = t]$
$ = 2 \int \frac { t ^ { 2 } + 1 } { t ^ { 4 } + 1 } d t$
On dividing numerator and denominator by $t^2$ , we get
$ I = 2 \int \frac { \left( 1 + \frac { 1 } { t ^ { 2 } } \right) } { \left( t ^ { 2 } + \frac { 1 } { t ^ { 2 } } \right) } d t $
$= 2 \int \frac { 1 + \frac { 1 } { t ^ { 2 } } } { t ^ { 2 } + \frac { 1 } { t ^ { 2 } } - 2 + 2 } d t$
$ = 2 \int \frac { \left( 1 + \frac { 1 } { t ^ { 2 } } \right) } { \left( t - \frac { 1 } { t } \right) ^ { 2 } + 2 } d t$
Again, put $ t-\frac { 1 } { t } = y \Rightarrow \left( 1 + \frac { 1 } { t ^ { 2 } } \right) d t = d y$
$ \therefore I = 2 \int \frac { d y } { y ^ { 2 } + ( \sqrt { 2 } ) ^ { 2 } } $
$ I = \frac { 2 } { \sqrt { 2 } } \tan ^ { - 1 } \frac { y } { \sqrt { 2 } } + C$$ \left[ \because \int \frac { d x } { \sqrt { x ^ { 2 } + a ^ { 2 } } } = \frac { 1 } { a } \tan ^ { - 1 } \left( \frac { x } { a } \right) + c \right]$
$ = \sqrt { 2 } \tan ^ { - 1 } \frac { \left( t - \frac { 1 } { t } \right) } { \sqrt { 2 } } + C \quad \left[ \text { put } y = t - \frac { 1 } { t } \right]$
$ = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { t ^ { 2 } - 1 } { \sqrt { 2 } t } \right) + C$
$ = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { \tan x - 1 } { \sqrt { 2 \tan x } } \right) + C$ $[Put \ t^2 = tan x]$
$I = \sqrt { 2 } \tan ^ { - 1 } \left( \frac { \tan x - 1 } { \sqrt { 2 \tan x } } \right) + C$
View full question & answer→