Definite Integration

$ \text { } \begin{aligned} & \int_0^{\frac{\pi}{4}} \frac{\sec x d x}{1+2 \sin ^2 x} \\ & =\int_0^{\pi / 4} \frac{\cos x d x}{\cos ^2 x\left(1+2 \sin ^2 x\right)} \end{aligned} $

$ \begin{aligned} &=\int_0^{\pi / 4} \frac{\cos x d x}{\left(1-\sin ^2 x\right)\left(1+2 \sin ^2 x\right)}\\ &\text { Let } \sin x=t \Rightarrow \cos x d x=d t\\ &\begin{aligned} & =\int_0^{1 / \sqrt{2}} \frac{d t}{\left(1-t^2\right)\left(1+2 t^2\right)} \\ & =\int_0^{1 / \sqrt{2}}\left(\frac{A}{1-t}+\frac{B}{1+t}+\frac{C t+D}{1+2 t^2}\right) d t \end{aligned} \end{aligned} $

$ \begin{aligned} & \begin{array}{r} =\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{6}\left(\frac{1}{1-t}\right) d t+\frac{1}{6} \int_0^{\frac{1}{\sqrt{2}}} \frac{d t}{t+1}+\frac{2}{3} \int_0^{\frac{1}{\sqrt{2}}} \frac{d t}{1+2 t^2} \\ \left(\because A=\frac{1}{6}=B \text { and } C=0, D=\frac{2}{3}\right) \end{array} \\ & \Rightarrow-\frac{1}{6}[\ln (1-t)]_0^{\frac{1}{\sqrt{2}}}+\frac{1}{6}[\ln (1+t)]_0^{\frac{1}{\sqrt{2}}} \\ & \quad+\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} t)\right] \\ & =-\frac{1}{6} \ln \left(1-\frac{1}{\sqrt{2}}\right)+\frac{1}{6}\left(\ln \left(1+\frac{1}{\sqrt{2}}\right)\right)+\frac{1}{3 \sqrt{2}} \tan ^{-1}(1) \\ & =\frac{1}{3} \ln (\sqrt{2}+1)+\frac{1}{3 \sqrt{2}} \cdot \frac{\pi}{4} \\ & =\frac{1}{3} \ln (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{n}{n} \cdot \sec ^2 \frac{n^2}{n^2}=\int_0^1 x \sec ^2 x^2 d x \\ & =\frac{1}{2} \int_0^1 2 x \sec ^2 x^2 d x \\ & \text { Let } x^2=z \Rightarrow 2 x d x=d z \\ & \qquad=\frac{1}{2} \int_0^1 \sec ^2 z d z=\frac{1}{2}(\tan z)_0^1 \\ & =\frac{1}{2} \tan 1 \end{aligned} \end{aligned} $

Given, $\int_2^5 \sqrt{\frac{5-x}{x-2}} d x$

Let $x$

$ \begin{aligned} x-2 & =z^2 \Rightarrow d x=2 z d z \\ & =\int_0^{\sqrt{3}} \frac{\sqrt{5-\left(z^2+2\right.}}{z} 2 z d z \\ & =2 \int_0^{\sqrt{3}} \sqrt{3-z^2} d z \\ & =2\left[\frac{z \sqrt{3-z^2}}{2}=\frac{3}{2} \sin ^{-1} \frac{z}{\sqrt{3}}\right]_0^{\sqrt{3}} \\ & =2\left[\frac{3}{2} \sin ^{-1}(1)\right]-3 \sin ^{-1}(1)=\frac{3 \pi}{2} \end{aligned} $

Given, $\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x d x$

Using Wallis formula,

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x, m, n \in z \\ & (m-1)(m-3) \ldots 2 \text { or } \cdot(n-1) \\ & =\frac{(n-3) \ldots 2 \text { or } 1}{(m+n)(m+n-2) \ldots 2 \text { or } 1} \times \lambda \end{aligned} $

where, $\lambda=\frac{\pi}{2}, m$ and $n$ both even.

$ \text { So, Let } \begin{aligned} \mathrm{I} & =\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x d x \\ & =\frac{5 \times 3 \times 1 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} \\ & =\frac{3 \pi}{512} \end{aligned} $

$ \begin{aligned} & \ \int_{1 / 2}^2\left|\log _{10} x\right| d x=\int_{1 / 2}^1\left(-\log _{10} x\right) d x \\ & +\int_1^2 \log _{10} x d x \\ & =-\int_{1 / 2}^1\left(\log _e x \cdot \log _{10} e\right) d x \\ & +\int_1^2\left(\log _e x \cdot \log _{10} e\right) d x \\ & =-\log _{10} e \int_{1 / 2}^1(\log x) d x+\log _{10} e \int_1^2(\log x) d x \\ & =-\log _{10} e[x(\log x-1)]_{\frac{1}{2}}^1 \\ & +\log _{10} e[x(\log x-1)]_1^2 \quad \text { (using by parts) } \\ & =-\log _{10} e\left[1(0-1)-\frac{1}{2}\left(\log \frac{1}{2}-1\right)\right] \\ & +\log _{10} e[2(\log 2-1)-1(0-1)] \\ & =\log _{10} e\left[\left(1+\frac{1}{2} \log \frac{1}{2}-\frac{1}{2}\right)+(2 \log 2-1)\right] \\ & =\left(\log _{10} e\right)\left(-\frac{1}{2} \log 2+2 \log 2-\frac{1}{2}\right) \\ & =\log _{10} e\left(\frac{3}{2} \log 2-\frac{1}{2}\right) \\ & =\left(\log _{10} e\right)\left(\frac{3}{2} \log _e 2-\frac{1}{2} \log _e e\right) \\ & =\frac{1}{2} \log _{10} e\left(\log _e 8-\log _e e\right) \\ & =\frac{1}{2}\left(\log _{10} e\right)\left(\log _e\left(\frac{8}{e}\right)\right)=\frac{1}{2} \log _{10}\left(\frac{8}{e}\right) \end{aligned} $

$ \begin{aligned} I & =\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x \\ I & =\int_0^{\pi / 2} \frac{\sin ^2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\ & =\int_0^{\pi / 2} \frac{\cos ^2 x}{\cos x+\sin x} d x \end{aligned} $

Adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 I=\int_0^{\pi / 2} \frac{1}{\sin x+\cos x} d x \\ & 2 I=\int_0^{\pi / 2} \frac{1}{\sqrt{2}\left(\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}\right)} d x \\ & 2 I=\int_0^{\pi / 2} \frac{1}{\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)^2} d x \\ & 2 I=\int_0^{\pi / 2} \frac{\sec \left(x-\frac{\pi}{4}\right)}{\sqrt{2}} d x \\ & 2 I=\frac{\ln \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right|_0^{\frac{\pi}{2}}}{\sqrt{2}} \\ & I=\frac{1}{2 \sqrt{2}}[\ln (\sqrt{2}+1)-\ln (\sqrt{2}-1)] \\ & =\frac{1}{2 \sqrt{2}} \ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\right) \\ & I=\frac{1}{\sqrt{2}} \ln (\sqrt{2}+1) \end{aligned} $

$ I=\int_0^{2 \pi}[|\sin x|+|\cos x|] d x$

$ \begin{aligned} & \text { Period of }|\sin x|+|\cos x| \text { is } \frac{\pi}{2} \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}}[|\sin x|+|\cos x|] d x \\ & \text { For } x \in\left(0, \frac{\pi}{2}\right), \sin x>0 \text { and } \cos x>0 \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}}[\sin x+\cos x] d x \\ & \Rightarrow \quad[\sin x+\cos x]=1 \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}} d x=\left.4 x\right|_0 ^{\frac{\pi}{2}}=2 \pi \end{aligned} $

Given, $f(x) \cdot f(-x)=9$

$ \begin{aligned} & f(-x)= \frac{9}{f(x)} \\ & I=\int_{-23}^{23} \frac{d x}{3+f(x)} \\ & I=\int_{-23}^{23} \frac{d x}{3+f(-x)} \\ &\left\{\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\} \\ & I=\int_{-23}^{23} \frac{d x}{3+\frac{9}{f(x)}} \quad \text { [from Eq. (i)] } \\ & I=\frac{1}{3} \int_{-23}^{23} \frac{f(x)}{3+f(x)} d x \end{aligned} $

Adding Eqs. (ii) and (iii), we get

$ \begin{aligned} 2 I & =\int_{-23}^{23}\left(\frac{1}{3+f(x)}+\frac{1}{3} \frac{f(x)}{3+f(x)}\right) d x \\ & =\int_{-23}^{23} \frac{3+f(x)}{3(3+f(x))} d x=\frac{1}{3} \int_{-23}^{23} d x \\ 2 I & =\frac{46}{3} \\ \Rightarrow I & =\frac{46}{6} \end{aligned} $

$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $

$3{x^2} - 5x + 2 = 0$

$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$

$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$

$ \Rightarrow (x - 1)(3x - 2) = 0$

$\therefore$ In 0 to ${2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2$

And in ${2 \over 3}$ to 1, $|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} } $

$ = \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} } $

Now, let $f(x) = - 3{x^2} + 7x - 1$

$ \Rightarrow f'(x) = - 6x + 7$

$ = - 6\left( {x - {7 \over 6}} \right)$

$f(0) = - 1$

$f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1$

$ = {7 \over 3} = 2.33$

$\therefore$ Range of $f(x) = \left[ { - 1,{7 \over 3}} \right]$

Integer value between $ - 1$ to ${7 \over 3} = - 1,0,1,2,{7 \over 3}$

When $f(x) = - 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = - 1$

$ \Rightarrow 3{x^2} - 7x = 0$

$ \Rightarrow x(3x - 7) = 0$

$ \Rightarrow x = 0,{7 \over 3}$ (not possible as ${7 \over 3} \notin (0,2)$)

When $f(x) = 0$

$ \Rightarrow - 3{x^2} + 7x - 1 = 0$

$ \Rightarrow 3{x^2} - 7x + 1 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}$

$ = {{7\, \pm \sqrt {37} } \over 6}$

$\therefore$ $x = {{7 - \sqrt {37} } \over 6}$

When $f(x) = 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = 1$

$ \Rightarrow 3{x^2} - 7x + 2 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}$

$ = {{7\, \pm \,\sqrt {25} } \over 6}$

$ = {{7\, \pm \,5} \over 6}$

$ \Rightarrow x = 2,\,{1 \over 3}$

When $f(x) = 2$

$ \Rightarrow - 3{x^2} + 7x - 1 = 2$

$ \Rightarrow 3{x^2} - 7x + 3 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}$

$ = {{7\, \pm \,\sqrt {13} } \over 6}$

$\therefore$ $x = {{7\, - \sqrt {13} } \over 6}$

$\therefore$ Possible values of $x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx} $

$ = \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } } $

$ = {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]$

$ = {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 8} \over 6}$

Now, $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

Let $f(x) = 3{x^2} - 3x + 3$

$f'(x) = 6x - 3 = 3(2x - 1)$

$f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}$

$f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)$

$ = {7 \over 3} = 2.3$

$f(1) = 3(1 - 1 + 1)$

$ = 3$

No integer values present in between ${7 \over 3}$ and 3.

$\therefore$ Possible value of $x = {2 \over 3},\,1$

So, integration don't break anywhere.

$\therefore$ $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

$ = \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx} $

$ = \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}} $

$\therefore$ Value of Integration

$ = {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}$

$ = {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 4} \over 6}$

$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $

$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $

Let $\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$

$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} $

$ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } $

$ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}$

$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $ ...... (i)

$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $

Substituting $t \to {1 \over p}$

$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $

$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $ ....... (ii)

By (i) + (ii)

$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $

$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$

$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$

${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $

$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $

$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt} $

$ = {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt} $

${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$

$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$

For $n = 5$

$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$

$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $

Differentiate on both side

${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$

$f(x) = {e^x}(2x - 1)$

$f'(x) = {e^x}(2) + {e^x}(2x - 1)$

$ = {e^x}(2x + 1)$

$x = - {1 \over 2}$

$f''(x) = {e^x}(2) + (2x + 1){e^x}$

$ = {e^x}(2x + 3)$

For $x = - {1 \over 2}\,\,f''(x) > 0$

$\Rightarrow$ Maxima

$\therefore$ Max. $ = {e^{ - {1 \over 2}}}( - 1 - 1)$

$\therefore$ $ - {2 \over {\sqrt e }}$

$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$

$\therefore$ $f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.$

$\therefore$ $f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$

and $\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $

$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$

$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$

$\therefore$ Only (S2) is correct

$\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } $

$ = \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } } $

$ = \left. {\left( {{{3{x^2}} \over 2} - {{2{x^3}} \over 3}} \right)} \right|_0^{3/2} + \left. {\left( {{{2{x^3}} \over 3} - {{3{x^2}} \over 2}} \right)} \right|_{3/2}^2 - {1 \over 2} + {1 \over 2}$

$ = \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)$

$ = {{19} \over {12}}$

$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$

Now,

$\because$ $\mathop {\lim }\limits_{x \to - 1} f(x)$ exist

$\therefore$ $\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$

$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$

$ \Rightarrow - a = 1 \Rightarrow a = - 1$

Now, $\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $

$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $

$ = 1 - 1 - 1 - 1 = - 2$

I comes out around 1.536 which is not satisfied by any given options.

$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $

${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $

${{\sin x} \over x}$ is decreasing in $\left( {{\pi \over 4},{\pi \over 3}} \right)$ so it attains maximum at $x = {\pi \over 4}$

$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $

$I > \sqrt 3 - {\pi \over 6}$

$\because$ f(x) is continuous at x = 4

$ \Rightarrow f({4^ - }) = f({4^ + })$

$ \Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt} $

$ = \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} } $

$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$

$ = 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$

$16 + 4b = 15$

$ \Rightarrow b = {{ - 1} \over 4}$

$ \Rightarrow f(x) = \left\{ {\matrix{ {\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr {{x^2} - {x \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {5 - |x - 3|} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {8 - x} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$

$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$

$f'(x) = 0 \Rightarrow x = {1 \over 8}$ have local minima

$\therefore$ (C) is only incorrect option.

$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $

$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $

$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $

$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$

$ = 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)$

$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $

$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $

$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} } $

$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)$

$f(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x$

$f'(x) = \cos e{c^2}x - \cos ec\,x\cot x$

$\left. {\matrix{ {f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \cr {f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \cr } } \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)$

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$

Let ${2^n} = t$ and if $n \to \infty $ then $t \to \infty $

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over t}\left( {\sum\limits_{r = 1}^{t - 1} {{1 \over {\sqrt {1 - {r \over t}} }}} } \right)$

$I = \int\limits_0^1 {{{dx} \over {\sqrt {1 - x} }} = \int\limits_0^1 {{{dx} \over {\sqrt x }}\,\,\,\,\,\,\,\,\,\left( {\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } } \right.} } $

$ = \left[ {2{x^{{1 \over 2}}}} \right]_0^1 = 2$

$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $

So,

$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $

$ = 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}$

$ = {{52} \over e}$

Case 1 :

Let $0 \le x < 1$

then $\left[ x \right] = 0$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 0 - x$

$ = 1 - x$

Case 2 :

Let $1 \le x < 2$

then $\left[ x \right] = 1$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 1$

Case 3 :

Let $2 \le x < 3$

then $\left[ x \right] = 2$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 2 - x$

$ = 3 - x$

Case 4 :

Let $3 \le x < 4$

then $\left[ x \right] = 3$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 3$

$\therefore$ $f(x) = \left\{ {\matrix{ {1 - x} & ; & {0 \le x < 1} \cr {x - 1} & ; & {1 \le x < 2} \cr {3 - x} & ; & {2 \le x < 3} \cr {x - 3} & ; & {3 \le x < 4} \cr } } \right.$

$\therefore$ $f(x)$ is periodic and period of $f(x) = 2$

And period of $\cos \pi x = {{2\pi } \over \pi } = 2$

$\therefore$ Period of $f(x)\cos \pi x = 2$

Now,

$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $

$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $

$\therefore$ $I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$

$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$

$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$

$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$

$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$

$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$

$ = {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]$

$ = {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$

$ = 4$

$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $

$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} $

$ = \int_a^b {f(x)dx} $

$a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0$

$b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1$

and ${r \over n} \to x$

$ = \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx} $

$ = \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx} $

$ = \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx} $

$ = \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx} $

$ = \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx} $

$ = \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1$

$ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]$

$ = - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)$

$ = + {3 \over 2}\log 3 - 2\log 2$

$ = \log \sqrt {{3^3}} - \log 4$

$ = \log {{\sqrt {{3^2} \times 3} } \over 4}$

$ = \log {{3\sqrt 3 } \over 4}$

Given,

$f(x) = x + \int_0^1 {(x - t)f(t)dt} $

$ = x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} } $

$ = x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt} $

Now,

let $A = 1 + \int_0^1 {f(t)dt} $

and $B = \int_0^1 {tf(t)dt} $

$\therefore$ $f(x) = Ax - B$

$ \Rightarrow f(t) = At - B$

So, $A = 1 + \int_0^1 {f(t)dt} $

$ = 1 + \int_0^1 {(At - B)dt} $

$ = 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1$

$ = 1 + {A \over 2} - B$

$ \Rightarrow {A \over 2} = 1 - B$ ...... (1)

$B = \int_0^1 {tf(t)dt} $

$ = \int_0^1 {t(At - B)dt} $

$ = \int_0^1 {(A{t^2} - Bt)dt} $

$ = \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1$

$ = {A \over 3} - {B \over 2}$

$ \Rightarrow {{3B} \over 2} = {A \over 3}$ ....... (2)

Solving equation (1) and (2) we get,

$A = {{18} \over {13}}$ and $B = {4 \over {13}}$

$\therefore$ $f(x) = {{18} \over {13}}x - {4 \over {13}}$

By checking all the options you can see when x = 6 we get

$y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}$

$ = {{108 - 4} \over {13}} = 8$

$\therefore$ Point (6, 8) lies on the curve.

Given,

$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $

Now,

$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} $

$ = \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} } $

$ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} $

$ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} $

[We know,

$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$]

$ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$

$ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$

$ = {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)$

$ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$

$ = {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$

$ = {8 \over 3} - {{2\pi } \over 4}$

$ = {8 \over 3} - {\pi \over 2}$

Now in R.H.S.,

$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} $

$ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } $

$ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$

$ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$

$ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$

$ = 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$

$ = {5 \over 6} - {\pi \over 4}$

Now,

$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} $

$ = \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$

$ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$

$ = 4 - {8 \over 6} - 2 + {1 \over 6}$

$ = 2 - {7 \over 6}$

$ = {5 \over 6}$

$\therefore$ $R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$

As $L.H.S. = R.H.S.$

$ \Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$

$ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$

$ \Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$

$ \Rightarrow I = 1 - {\pi \over 4}$

From option (c)

$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy} $

$ = \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy} $

$ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$

$ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$

$ = 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$

$ = 1 - {\pi \over 4} = I$

$\therefore$ Option (c) is the right answer.

Note :

There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.

We know,

$\left[ {{x \over 2}} \right]$ is discontinuous at 1, 2, 3, 4 ........

$\therefore$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts

(1) 0 to 2

(2) 2 to 4

(3) 4 to 5

$\therefore$ $\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $

$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $

$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $

$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $

$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$

$ = 0 - 0 + 0$

$ = 0$

$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $

$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$

$x = 0$, or $({x^2} - 4)({x^2} - 1) = 0$

$x = 0,$ $x = \pm 2,\, \pm 1$

Now, $f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\left( {{e^{{x^2}}} + 2} \right)}}$

f'(x) changes sign from positive to negative at x = $-$1, 1 So, number of local maximum points = 2

f'(x) changes sign from negative to positive at x = $-$2, 0, 2 So, number of local minimum points = 3

$\therefore$ m = 2, n = 3

$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $

$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$

$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$

$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$

Put $\cos x = {1 \over {\sqrt 3 }}$,

$\therefore$ $f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2 $

${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}$

$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx} $, let ${1 \over x} = t$

${{ - 1} \over {{x^2}}}dx = dt$

$ = \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} } $

$ = \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} } $

$ = {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$

$ = \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $

$ = \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} } $

$ = - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$

$ = 1 + 6\log {6 \over 7}$

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $ ..... (i)

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $ ..... (ii)

$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $

$2I =2 \int\limits_0^2 {({x^3} + x)dx} $

$I = \int\limits_0^2 {({x^3} + x)dx} $

$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$

$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$

$ = 4 + 2 = 6$

${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $

$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $

$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$

So, ${b_3} - {b_2}$, ${b_4} - {b_3}$, ${b_5} - {b_4}$ are in H.P.

$ \Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$ are in A.P. with common difference $-$2.

$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} $

Let $\cos x = t$

$\sin xdx = dt$

$ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} $

$I = \int\limits_{ - 1}^1 {{{{e^t}} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}dt} $ ...... (i)

$I = \int\limits_{ - 1}^1 {{{{e^{ - t}}} \over {\left( {1 + {t^2}} \right)\left( {{e^{ - t}} + {e^t}} \right)}}dt} $ .... (ii)

Adding (i) and (ii)

$2I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}} $

$2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1$

$2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)$

$2I = {\pi \over 2}$

$I = {\pi \over 4}$

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $ ...... (i)

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $ ..... (ii)

(i) and (ii)

From equation (i) & (ii)

$2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} } $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} } $

when x = 0, t = 0

Now, $\tan 2x = t$

when $x = {\pi \over 4},\,\,t \to \infty $

$2{\sec ^2}2x\,dx = dt$

$\therefore$ $I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty } $

$ = 2{\pi \over 2} = \pi $

$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}} $

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left[ {1 + {{\left( {{r \over n}} \right)}^2}} \right]\left[ {1 + \left( {{r \over n}} \right)} \right]}}} $

$ = \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx} $

$ = {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx} $

$ = {1 \over 2}\left[ {\ln (1 + x) - {1 \over 2}\ln \left( {1 + {x^2}} \right) + {{\tan }^{ - 1}}x} \right]_0^1$

$ = {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2$