Definite Integration

Given differential equation

$f'(x) = {{f(x)} \over {{b^2} + {x^2}}}$

$ \Rightarrow \int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over {{b^2} + {x^2}}}} } $

$ \Rightarrow {\log _e}|f(x)| = {1 \over b}{\tan ^{ - 1}}\left( {{x \over b}} \right) + c$

$ \because $ f(0) = 1, so c = 0

$ \therefore $ $|f(x)| = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

$ \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$ or $ - {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

$ \because $ $f(0) = 1 \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

and $f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}$

$ \therefore $ $f'(x) > 0\forall x \in R$ and $b \in {R_0}$.

Therefore, f(x) is an increasing function $\forall b \in {R_0}$.

and f(x) f($-$x)

$ = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}\,.\,{e^{ - {1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}} = 1$

$ \because $ $\cos x = 1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} - {{{x^6}} \over {6!}} + ...$

and $\sin x = x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - {{{x^7}} \over {7!}} + ....$

$ \therefore $ $\int_0^1 {x\cos xdx} \ge \int_0^1 {\left( {x - {{{x^3}} \over {2!}}} \right)} \,dx$

$ = \left[ {{{{x^2}} \over 2} - {{{x^4}} \over 8}} \right]_0^1 = {1 \over 2} - {1 \over 8} = {3 \over 8}$

$ \Rightarrow \int_0^1 {x\cos xdx\, \ge \,{3 \over 8}} $

and, $\int_0^1 {x\sin dx \ge \int_0^1 {\left( {{x^2} - {{{x^4}} \over 6}} \right)} \,dx} $

$\left[ {{{{x^3}} \over 3} - {{{x^5}} \over {30}}} \right]_0^1 = {1 \over 3} - {1 \over {30}} = {9 \over {30}} = {3 \over {10}}$

$ \Rightarrow \int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}} $

and, $\int_0^1 {{x^2}\cos xdx\, \ge \,\int_0^1 {\left( {{x^3} - {{{x^5}} \over 2}} \right)\,dx} } $

$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {12}}} \right]_0^1 = {1 \over 4} - {1 \over {12}} = {2 \over {12}} = {1 \over 6}$

$ \therefore $ $\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 6}} $

and, $\int_0^1 {{x^2}\sin xdx\, \ge \int_0^1 {\left( {{x^3} - {{{x^5}} \over 6}} \right)\,dx} } $

$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {36}}} \right]_0^1 = {1 \over 4} - {1 \over {36}} = {8 \over {36}} = {2 \over 9}$

$ \therefore $ $\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}} $

$ \begin{aligned} &\text { Given that, }\\ &F(x)=\left\lvert\, \begin{array}{cc} 1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} \\ 3+4 \sin x & 3 / 2 \\ 1+\sin x & \frac{1}{2} \sin x \end{array}\right. \end{aligned} $

$\left. \matrix{ {{ - 2 + \sin 3x} \over 3} \hfill \cr {4 \over \matrix{ 3 \hfill \cr {1 \over 3} \hfill \cr} }\sin x\,\,\,\,\,\,\,\,\, \hfill \cr} \right|$

$ \begin{aligned} &\begin{aligned} & C_1 \rightarrow C_1-2 C_2-3 C_3 \\ & f(x)=\left|\begin{array}{ccc} \sin x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin x \\ 0 & \frac{1}{2} \sin x & \frac{1}{3} \end{array}\right| \end{aligned}\\ &\text { On expanding, we get }\\ &\begin{aligned} f(x) & =\sin x\left|\begin{array}{cc} \frac{3}{2} & \frac{4}{3} \sin x \\ \frac{1}{2} \sin x & \frac{1}{3} \end{array}\right|+0+0 \\ & =\sin x\left(\frac{3}{2} \times \frac{1}{3}-\frac{4}{3} \sin x \times \frac{1}{2} \sin x\right) \\ & =\sin x\left(\frac{1}{2}-\frac{2}{3} \sin ^2 x\right) \\ f(x) & =\frac{1}{2} \sin x-\frac{2}{3} \sin ^3 x \\ f(x) & =\frac{1}{6}\left(3 \sin x-4 \sin ^3 x\right) \end{aligned} \end{aligned} $

$ f(x)=\frac{\sin 3 x}{6}\left[\because \sin 3 x=3 \sin x-4 \sin ^3 x\right] $

Differentiating on both sides w.r.t. ' $x$ ', we get

$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left(\frac{\sin 3 x}{6}\right)=\frac{1}{6} \cdot \cos 3 x \cdot 3 \\ & f^{\prime}(x)=\frac{\cos 3 x}{2} \end{aligned} $

$ \begin{align*}Let\,\, I & =\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x \\ I & =\int_0^{\pi / 2}\left(\frac{\sin 3 x}{6}+\frac{\cos 3 x}{2}\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

$ \begin{align*} &or\,\, I=\int_0^{\pi / 2}\left(\frac{\sin 3\left(\frac{\pi}{2}-x\right)}{6}+\frac{\cos 3\left(\frac{\pi}{2}-x\right)}{2}\right) d x \\ & =\int_0^{\pi / 2}\left(\frac{\sin \left(\frac{3 \pi}{2}-3 x\right)}{6}+\frac{\cos \left(\frac{3 \pi}{2}-3 x\right)}{2}\right) d x \\ & I=\int_0^{\pi / 2}\left(-\frac{\cos 3 x}{6}-\frac{\sin 3 x}{2}\right) d x \quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

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

$ 2 I=\int_0^{\pi / 2}\left(\frac{\cos 3 x}{2}-\frac{\cos 3 x}{6}+\frac{\sin 3 x}{6}-\frac{\sin 3 x}{2}\right)dx $

$ \begin{aligned} & 2 I=\frac{1}{3} \int_0^{\pi / 2}(\cos 3 x-\sin 3 x) d x \\ \text { or } & 2 I=\frac{1}{3}\left[\frac{\sin 3 x}{3}+\frac{\cos 3 x}{3}\right]_0^{\pi / 2} \\ = & \frac{1}{9}\left[\begin{array}{r} \sin 3 \times \frac{\pi}{2}+\cos 3 \times \frac{\pi}{2}-\sin 3 \\ \times 0-\cos 3 \times 0 \end{array}\right] \\ = & \frac{1}{9}[-1+0-0-1]=-\frac{2}{9} \text { or } I=-\frac{1}{9} \end{aligned} $

Given that,

$ \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{n} \sin ^{-1} \frac{1}{n}+\frac{2}{n} \sin ^{-1} \frac{2}{n}+\ldots+\frac{\pi}{2}\right] $

The above expression may be written as

$ =\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n \frac{r}{n} \sin ^{-1} \frac{r}{n}}{n}=\int_0^1{\underset{\mathrm{II}}{x}}^x \sin _{\mathrm{I}}^{-1} x d x $

Solving integration by parts method, we get

$ \begin{aligned} =\sin ^{-1} x & \int_0^1 x d x \\ & -\int_0^1\left(\frac{d}{d x}\left(\sin ^{-1} x\right) \int x d x\right) d x \end{aligned} $

$ \begin{aligned} & =\left[\sin ^{-1} x \cdot \frac{x^2}{2}\right]_0^1-\int_0^1 \frac{1}{\sqrt{1-x^2}} \times \frac{x^2}{2} d x \\ & =\left[\sin ^{-1} x \cdot \frac{x^2}{2}\right]_0^1-\frac{1}{2} \int_0^1 \frac{x^2-1+1}{\sqrt{1-x^2}} d x \\ & =\left[\sin ^{-1}(1) \cdot \frac{1}{2}-\sin ^{-1}(0) \cdot \frac{0^2}{2}\right]-\frac{1}{2} \\ & \int_0^1 \frac{-\left(1-x^2\right)}{\sqrt{1-x^2}} d x-\frac{1}{2} \int_0^1 \frac{d x}{\sqrt{1-x^2}} \\ & =\left[\frac{\pi}{2} \times \frac{1}{2}-0\right]+\frac{1}{2} \int_0^1 \sqrt{1-x^2} d x -\frac{1}{2} \int_0^1 \frac{d x}{\sqrt{1-x^2}} \end{aligned} $

$ \begin{aligned} & =\frac{\pi}{4}+\frac{1}{2}\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^1 \\ & -\frac{1}{2}\left[\sin ^{-1} x\right]_0^1 \\ & {\left[\because \int \sqrt{a^2-x^2} d x=\frac{1}{2 a} \sqrt{a^2-x^2}\right.} \\ & +\frac{1}{2 a} \sin ^{-1}\left(\frac{x}{a}\right) \text { and } \\ & \left.\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)\right] \\ & =\frac{\pi}{4}+\frac{1}{2} \\ & {\left[\frac{1}{2} \sqrt{1-1^2}+\frac{1}{2} \sin ^{-1}(1)-\frac{0}{2} \sqrt{1-0^2}\right.} \\ & \left.-\frac{0}{2} \sin ^{-1}(0) \frac{-1}{2}\left[\sin ^{-1}(2)-\sin ^{-1}(0)\right]\right] \\ & =\frac{\pi}{4}+\frac{1}{2}\left[0+\frac{1}{2} \frac{\pi}{2}-0-0\right]-\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\ & =\frac{\pi}{4}+\frac{1}{2}\left[\frac{\pi}{4}-0\right]-\frac{\pi}{4}=\frac{\pi}{4}+\frac{\pi}{8}-\frac{\pi}{4}=\frac{\pi}{8} \end{aligned} $

Given,

$ \begin{aligned} f(x) & =\frac{1}{x^3} \int_5^x\left(2 u^2-u f^{\prime}(u) d u\right. \\ \Rightarrow \quad x^3 f(x) & =\int_5^x\left(2 u^2-u f^{\prime}(u)\right) d u \end{aligned} $

On differentiating w.r.t ' $x$ ',

$ \begin{aligned} & x^3 f^{\prime}(x)+3 x^2 f(x)=2 x^2-x f^{\prime}(x) \\ & f^{\prime}(x)=\frac{2 x^2-3 x^2 f(x)}{x^3+x} \\ & f^{\prime}(5)=\frac{2(5)^2-3(5)^2 f(5)}{5^3+5} \\ & f^{\prime}(5)=\frac{50-0}{130}=\frac{50}{130} \quad[\because f(5)=0] \\ & f^{\prime}(5)=\frac{5}{13} \end{aligned} $

We have,

$ \begin{aligned} A & =\int_{-a}^a f(x) d x \\ & =\int_0^a(f(x)+f(-x)) d x \end{aligned} $

A is true,

$ R=\int_a^b f(x) d x=\int_{g(a)}^{g(b)} f\left(g(u) g^{\prime}(u) d u\right. $

Put, $x=g(u), d x=g(u) d u$

at $x=a=g(u)$ and $x=b$

$ \begin{aligned} & u=g^{-1}(b) \Rightarrow u=g^{-1}(a) \\ & \therefore \int_a^b f(x) d x=\int_{g^{-1}(a)}^{g^{-1}(b)} f\left(g(x) g^{\prime}(x) d x\right. \end{aligned} $

$\therefore \mathrm{R}$ is false.

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \cos x+\cos 2 x+\ldots .+\cos n x=\frac{A(x)}{2 \sin x / 2} \\ & \therefore \quad A(x)=2 \cos \left(x+\frac{(n-1)}{2} x\right) \sin \frac{n x}{2} \\ & A(x)=2 \cos \left(\frac{n+1}{2}\right) x \sin \frac{n x}{2} \\ & A(x)=\sin \left(\frac{2 n+1}{2}\right) x-\sin \frac{x}{2} \\ & \int_0^\pi A(x) d x=\int_0^\pi\left(\sin \left(\frac{2 n+1}{2}\right) x-\sin \frac{x}{2}\right) d x \\ & =\left[-\frac{2}{2 n+1} \cos \left(\frac{2 n+1}{2}\right) x+2 \cos \frac{x}{2}\right]_0^\pi \\ & =\left(\frac{-2}{2 n+1} \cos \left(\frac{2 n+1}{2}\right) \pi+2 \cos \frac{\pi}{2}\right) -\left(\frac{-2}{2 n+1} \cos 0+2 \cos 0\right) \\ & =0-\left[\frac{-2}{2 n+1}+2\right]=-\frac{4 n}{2 n+1} \end{aligned} \end{aligned} $

We have,

$ \lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right] $

Here, $r$ th term $=\sin \left(\frac{r \pi}{2 n}\right)$

∴ The given limit, $\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \sum_{r=1}^n \sin \left(\frac{r \pi}{2 n}\right)$

$ \begin{gathered} \text { Put } \frac{r \pi}{2 n} \rightarrow x, \frac{\pi}{2 \pi} \rightarrow d x, \quad \lim _{n \rightarrow \infty} \Sigma \rightarrow \int \\ x_{\min }=\lim _{n \rightarrow \infty} \frac{r_{\min }}{2 n / \pi}=0 \end{gathered} $

$ \begin{aligned} x_{\max } & =\lim _{n \rightarrow \infty} \frac{r_{\max }}{2 n / \pi}=\frac{\pi}{2} \\ & =\int_0^{\pi / 2} \sin x d x \\ & =(-\cos x)_0^{\pi / 2}=-(\cos x)_0^{\pi / 2} \\ & =-\left[\cos \frac{\pi}{2}-\cos 0\right] \\ & =-[0-1]=1 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \text { ee, } \int_0^{\pi / 2} \frac{d x}{4+5 \sin x} \\ & =\int_0^{\pi / 2} \frac{d x}{4+5\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\int_0^{\pi / 2} \frac{\left(1+\tan ^2 \frac{x}{2}\right)}{4+4 \tan ^2 \frac{x}{2}+10 \tan \frac{x}{2}} d x \\ & =\frac{1}{4} \int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2}}{\tan ^2 \frac{x}{2}+\frac{5}{2} \tan \frac{x}{2}+1} d x \\ & \text { let } \tan \frac{x}{2}=t \\ & \Rightarrow \quad \sec ^2 \frac{x}{2}\left(\frac{1}{2}\right) d x=d t \\ & \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t \text { at } x=0, t=0 \\ & \text { and } x=\frac{\pi}{2}, t=1=\frac{2}{4} \int_0^1-\frac{1}{t^2+\frac{5}{2} t+1} d t \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \int_0^1 \frac{d t}{t^2+\frac{5}{2} t+1+\frac{25}{16}-\frac{25}{16}} \\ & =\frac{1}{2} \int_0^1 \frac{d t}{\left(t+\frac{5}{4}\right)^2-\frac{9}{16}} \\ & =\frac{1}{2} \int_0^1 \frac{d t}{\left(t+\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2} \\ & =\frac{1}{2}\left[\frac{1}{2 \times\left(\frac{3}{4}\right)} \log \left|\frac{t+\frac{5}{4}-\frac{3}{4}}{t+\frac{5}{4}+\frac{3}{4}}\right|\right]_0^1 \\ & =\frac{1}{3}\left[\left.\log \left(\frac{t+\frac{1}{2}}{t+2}\right)\right|_0 ^1\right. \\ & =\frac{1}{3}\left[\log \left(\frac{3 / 2}{3}\right)-\log \left(\frac{1 / 2}{2}\right)\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{3}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{4}\right)\right]=\frac{1}{3}\left[\log \left(\frac{\frac{1}{2}}{\frac{1}{4}}\right)\right] \\ & =\frac{1}{3} \log 2 \end{aligned} $

$ \begin{aligned} &\text { Let }\\ &\begin{aligned} & L=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}} \\ & \therefore \quad \log L=\lim _{n \rightarrow \infty} \frac{1}{n} \\ & \quad\left[\log \left(1+\frac{1}{n^2}\right)+\log \left(1+\frac{2^2}{n^2}\right)\right.\left.+\ldots . \log \left(1+\frac{n^2}{n^2}\right)\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\left(\frac{r}{n}\right)^2\right) \\ & =\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[x \log \left(1+x^2\right)\right]_0^1-\int_0^1 x \cdot \frac{1}{1+x^2} \cdot 2 x d x \\ & =\log 2-2 \int_0^1 \frac{x^2}{1+x^2} d x \\ & =\log 2-2 \int_0^1\left(1-\frac{1}{1+x^2}\right) d x \\ & =\log 2-2\left[x-\tan ^{-1} x\right]_0^1 \end{aligned} \end{aligned} $

$ \begin{aligned} & =\log 2-2\left[1-\frac{\pi}{4}\right] \\ & =\log 2-2+\frac{\pi}{2}=\log 2-2+\frac{\pi}{2} \\ & \therefore L=e^{e \log 2} \cdot e^{\frac{\pi}{2}-2}=2 e^{\frac{\pi-4}{2}} \end{aligned} $

$ \begin{aligned} &\text { Let }\\ &\begin{aligned} I & =\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} \ldots(\mathrm{i}) \\ & =\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}-x-\frac{3 \pi}{8}\right)}} \end{aligned} \end{aligned} $

$ \begin{aligned} & {\left[\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] } \\ = & \int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{8}-x\right)}} \\ = & \int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{-\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} \\ = & \int_{\pi / 4}^{\pi / 2} \frac{3 e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)} d x}{e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)+1}} \quad \ldots \text { (i) } \end{aligned} $

$ \begin{aligned} &\text { On adding Eq. (i) and Eq. (ii), we get }\\ &\begin{aligned} 2 I & =3 \int_{\pi / 4}^{\pi / 2} 1 d x=3[x]_{\pi / 4}^{\pi / 2} \\ & =3\left[\frac{\pi}{2}-\frac{\pi}{4}\right]=\frac{3 \pi}{4} \\ \therefore \quad I & =\frac{3 \pi}{8} \end{aligned} \end{aligned} $

$I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}} $

Let x + $\alpha $ = t and dx = dt

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

$ \Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}$

$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)$

$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)$

$ \Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}$

$4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha $

$ \Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0$

$ \Rightarrow {\alpha ^2} + \alpha - 2 = 0$

$ \Rightarrow (\alpha + 2)(\alpha - 1) = 0$

$\alpha $ = 1 or $\alpha $ = -2

$\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx} $

$ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} $

$ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)$

mn = ${1 \over 2}x - 2 = - 1$

Given $\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)$

$ \Rightarrow g(x) = $ ${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$

$ \therefore $ $\mathop {\lim }\limits_{x \to 2} g(x) = $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$

At x = 2 this limit is in ${0 \over 0}$ form.

So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.

$ \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18$

$\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx} $

$ = \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx} $

Let tan x = t, sec² x dx = dt

$ = \int {{{dt} \over {{t^{4/3}}}}} $

$I = - 3\left( {{t^{ - 1/3}}} \right)$

$ \Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}$

$ \Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right)$ = 3^7/6 - 3^5/6

$I = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx$ .... (i)

$ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx$

$ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx$

By (i) + (ii)

$ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} $

$ \Rightarrow 2I = - (x)_0^{2\pi }$

$ \Rightarrow $ I = –$\pi $

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

$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3}}}}}\,\,\,{r \over n} \to x\,and\,{1 \over n} \to dx\,} $

$ \Rightarrow \int\limits_0^1 {{{(1 + x)}^{{1 \over 3}}}dx} $

$ \Rightarrow \left[ {{3 \over 4}{{(1 + x)}^{{4 \over 3}}}} \right]_0^1 = {3 \over 4}{(2)^{{4 \over 3}}} - {3 \over 4}$

$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$

This is ${0 \over 0}$ form so we use L – Hospital Rule.

= $\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$

= ${2f\left( 2 \right).f'\left( 2 \right)}$

= 12f'(2)

Note : Newton - Leibnitz rule of differentiation under integration

${d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du$

= $f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)$

I = $\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $

Let x² = t

$ \Rightarrow $ 2xdx = dt

At x = 0, t = 0

At x = 1, t = 1

I = ${1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt} $

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

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

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

= ${1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt} $

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

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

[ As ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $ ]

= $2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $

= $\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $

Using integration by parts rule

= $\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1$$ - \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt$

= ${\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1$

= ${\pi \over 4} - {1 \over 2}{\log _e}2$

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

$ \Rightarrow $ I = $\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx$ ......(2)

Adding those two

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

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

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

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

= $\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}$

= $\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)$

$ \therefore $ 2I = $\left( {{\pi \over 2} - {1 \over 2}} \right)$

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

$f(x) = \int\limits_0^x {g(t)dt} $

$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$

Put t = -v

= $ - \int\limits_0^x {g\left( { - v} \right)} dv$

= $ - \int\limits_0^x {g\left( v \right)} d(v)$ [ as g(v) is an even function.]

= - f(x)

$ \Rightarrow $ f(-x) = -f(x)

$ \therefore $ f(x) is an odd function.

Given ƒ(x + 5) = g(x)

$ \therefore $ g(- x) = ƒ(- x + 5)

$ \Rightarrow $ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]

Replacing x by x + 5, we get

f(x) = - g(x + 5) ......(1)

Now

$ \int\limits_0^x {f(t)dt} $

= $ - \int\limits_0^x {g\left( {t + 5} \right)} dt$

= $ - \int\limits_5^{x + 5} {g\left( t \right)} dt$

= $\int\limits_{x + 5}^5 {g(t)dt} $

$g\left( {f\left( x \right)} \right)$ = $\ln \left( {f\left( x \right)} \right)$ = $\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)$

$ \therefore $ I = $\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} $

( Using property $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx$ )

I = $\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} $

= $\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} $

= 0 = log_e1

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

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2$

$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$

Let  ${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$

$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $

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

${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $

${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $

${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $

${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $

$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $

I $=$ $\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $

${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $   Put tan⁵x $=$ t

${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$

I $=$ $\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx} $

$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$   as   $\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0} $

${\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} $

$ = \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} } $

$\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}} $

$\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)$

$ - 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}$

${{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}$

${{ - 9} \over {20}} + {{3\pi } \over 5}$

$\int\limits_0^x \, $f(t) dt = x² + $\int\limits_x^1 \, $ t²f(t) dt           f '$\left( {{1 \over 2}} \right)$ = ?

Differentiate w.r.t. 'x'

f(x) = 2x + 0 $-$ x² f(x)

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

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

f '$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$

Let f(x) = x²(x² $-$ 2)


As long as f(x) lie below the x-axis, define integral will remain negative,
so correct value of (a, b) is ($-$ $\sqrt 2 $, $\sqrt 2 $) for minimum of I

$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$

$ \Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)$

$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}$

$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} \right) = 1 - {1 \over {\sqrt 2 }}$

$ \Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }}$

$ \Rightarrow \,\,k = 2$

$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$

$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$

$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$

$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$  $ \Rightarrow f'\left( x \right) = 0$

$ \Rightarrow f\left( x \right) = $ constant

as  $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$

$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$

$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$

The period of $\left| {\cos x} \right|$ = ${\pi \over 2}$

$ \therefore $  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$

as in the range 0 to ${\pi \over 2}$ $\left| {\cos x} \right|$ is positive.

So,     $\left| {\cos x} \right|$ = $cosx$

$ \therefore $  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $

= 2$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$

I = ${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$

I = ${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$

I = ${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$

= ${1 \over 2}\left( {{8 \over 3}} \right)$

= ${4 \over 3}$

f(x) = $\int_0^x {t(\sin x - \sin t).dt} $

= sin x$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $

= ${{{x^2}} \over 2}$ sin x +$\left[ {t\cos t} \right]_0^x$ + sin x

$ \Rightarrow $f(x) = ${{{x^2}} \over 2}$ sinx + xcosx + sinx

f'(x) = ${{{x^2}} \over 2}$ cosx + 2cos x

f''(x) = x cos x $-$ ${{{x^2}} \over 2}$ sin x $-$ 2sin x

f'''(x) = cos x $-$ 2x sin x $-$ ${{{x^2}} \over 2}$ cos x $-$ 2cos x

$\therefore\,\,\,$ f'''(x) + f'(x) = cos x $-$ 2x sin x

As we know,

$\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$

Let $I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$

$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $

$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$

$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$

$\therefore\,\,\,$ $2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$

$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$

$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$

$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$ [ as sin² x is an even function ]

$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$

[ as $\,\,\,\,$ $\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $

So, $\,\,\,$ $I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$

$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$

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

$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$

$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$

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

$ \Rightarrow \,\,\,\,I = {\pi \over 4}$

Given,

${I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;$

${I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx$ and

${I_3} = \int_0^1 {{e^{ - {x^3}}}} dx$

For x $ \in $ (0, 1)

$ \Rightarrow $  x > x² or $-$ x < $-$ x²

and x² > x³ or $-$ x² < $-$ x³

$ \therefore $ ${e^{ - {x^2}}}$ < ${e^{ - {x^3}}}$ and ${e^{ - x}}$ < ${e^{ - {x^2}}}$

$ \Rightarrow $ ${e^{ - x}} < {e^{ - {x^2}}} < {e^{ - {x^3}}}$

$ \Rightarrow $ ${e^{ - {x^3}}} > {e^{ - {x^2}}} > {e^{ - x}}$

$ \Rightarrow $ ${I_3} > {I_2} > {I_1}$

Let  $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} $

also let K = ${x \over {1 + \sin x}}$

Multiplying numerator and denominator by
(1 $-$ sin x) we get;

$K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}$

= x(1 $-$ sin x) sec²x

= x sec²x $-$ x sin x sec²x = x sec²x $-$ x tan x sec x

Now, $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} } $

$ \Rightarrow $ I = $\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

- $\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}} $

$ \Rightarrow $ I = $\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$ - $\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$

$ - \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

$ \Rightarrow $ I = $ - {{3\pi } \over 4} - {\pi \over 4}$$ + {{3\pi } \over 4}\left( {\sqrt 2 } \right)$$ + {\pi \over 4}\left( {\sqrt 2 } \right)$

$ - \log \left( {\sqrt 2 + 1} \right)$$ + \log \left( {\sqrt 2 + 1} \right)$

$ \Rightarrow $ I = -$\pi $ + ${\sqrt 2 \pi }$

$ \Rightarrow $ I = $\pi \left( {\sqrt 2 - 1} \right)$

Let

I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$ ......(1)

$ \Rightarrow $ I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx$

[ As $\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$ ]

     = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx$

     = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$ .......(2)

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

2I = 2$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx$

2I = 4$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$

I = 2$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$

  = ${{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}$ [ Using Gamma function ]

  = ${{3\pi } \over 8}$

$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$

$ \Rightarrow $   $\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$

        $ = {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$

        $ = {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$

$ \Rightarrow $  ${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$

$ \Rightarrow $  2a² + 3a $-$ 119 $=$ 0

$ \Rightarrow $   2a² + 17a $-$ 14a $-$ 119 $=$ 0

$ \Rightarrow $   (a $-$ 7) (2a + 17) $=$ 0

$ \Rightarrow $   a $=$ 7, $-$ ${{17} \over 2}$

Given, I = $\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} $

= $\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} $

Let x $-$ 1 = $\sqrt 3 $ tan$\theta $

$ \Rightarrow $$\,\,\,$ dx = $\sqrt 3 $ sec²$\theta $ d$\theta $

When x = 1, then $\theta $ = 0

and when x = 2, $\theta $ = ${\pi \over 6}$

I = $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}} $

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}} $

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}} $

= ${1 \over 3}$ $\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}} $

= ${1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta } $

= ${1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}$

= ${1 \over 3} \times {1 \over 2}$

= ${1 \over 6}$

$\therefore\,\,\,$ According to the question,

${k \over {k + 5}}$ = ${1 \over 6}$

$ \Rightarrow $$\,\,\,$ 6k = k + 5

$ \Rightarrow $$\,\,\,$ k = 1

tan x  +  cot x

= ${{\sin x} \over {\cos x}}$ + ${{\cos x} \over {\sin x}}$

= ${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$

= ${1 \over {\sin x\,\,\cos x}}$

= ${2 \over {\sin 2x}}$

$\therefore\,\,\,$ $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx$

Let sin 2x = t

$ \Rightarrow $$\,\,\,$ 2 cos2x dx = dt

At    x = ${{\pi \over {12}}}$, t = sin ${{\pi \over {6}}}$ = ${1 \over 2}$

At   x = ${{\pi \over 4}}$, t = sin ${{\pi \over 2}}$ = 1.

= ${1 \over 2}$ $\int\limits_{{1 \over 2}}^1 {{t^3}.dt} $

= ${{1 \over 2}}$ $\left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1$

= ${{1 \over 2}}$ $ \times $ ${{1 \over 4}}$ $ \times $ [ 1⁴ $-$ (${{1 \over 2}}$)⁴]

= ${{1 \over 8}}$ $ \times $ ${{15 \over 16}}$ = ${{15 \over 128}}$

$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $

= $\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $

= ${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$

${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

= $\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$

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

= $\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)$

= 2

$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $

Clearly, $I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x{{(x + 1)}^2}}}dx} } $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {(k + 1)\int_k^{k + 1} {{1 \over {{{(x + 1)}^2}}}dx} } $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {( - (k + 1))\left[ {{1 \over {k + 2}} - {1 \over {k + 1}}} \right]} $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {{1 \over {k + 2}}} $

$ \Rightarrow I > {1 \over 3} + ... + {1 \over {100}} > {{98} \over {100}}$

$ \Rightarrow I > {{49} \over {50}}$

Also, $I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $

$ = \sum\limits_{k = 1}^{98} {[{{\log }_e}(k + 1) - {{\log }_e}k]} $

$I < {\log _e}99$

$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$

Differentiate w.r. to x.

$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $

$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $

$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $

Diff. again w.r.t to x

$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$
$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$

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

${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$

$\ln \left( {y{x^3}} \right) = - {1 \over x}$

$y{x^3} = - {e^{ - 1/x}}$

$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$

or   $y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$

Let   I   =   $\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $

  =   $\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$

Using,

$\int\limits_a^b {f\left( {a + b - x} \right)dx\,} $ = $\,\,\int\limits_a^b {f(x)\,\,dx} $

I   =   $\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$

Adding (1) and (2)

2I   =   $\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$

$ \Rightarrow $$\,\,\,$ 2I   =  $\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$ = 6

=   I = 3