Definite Integration

$\begin{aligned} & \text { (c) Let } I=\int_{-1 / 2}^{1 / 2}\left\{[x]+\log \left(\frac{1+x}{1-x}\right) d x\right\} \\ & \because \int_{-a}^a f(x) d x=\left\{\begin{array}{cr} 2 \int_0^a f(x) d x, & f(-x)=f(x) \\ 0, & f(-x)=-f(x) \end{array}\right. \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) d x \end{aligned}$

Let $f(x)=\log \left(\frac{1+x}{1-x}\right)$

$\begin{aligned} & f(-x)=\log \left(\frac{1-x}{1+x}\right)=-f(x) \quad \quad \text { [odd function] } \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+0=\int_{-1 / 2}^0[x] d x+\int_0^{1 / 2}[x] d x \\ & =\int_{-1 / 2}^0(-1) d x+\int_0^{1 / 2} 0 d x \\ & =-(x)_{-1 / 2}^0+0=-\left(0+\frac{1}{2}\right)=-\frac{1}{2} \quad \end{aligned}$

Let $I = \int_0^1 {{{\log (1 + x)} \over {1 + {x^2}}}dx = \int_0^{\pi /4} {{{\log (1 + \tan \theta )} \over {(1 + {{\tan }^2}\theta )}}{{\sec }^2}\theta d\theta } } $

Let $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta d\theta $

$ = \int_0^{\pi /4} {\log (1 + \tan \theta )d\theta = \int_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]d\theta } } $

$ = \int_0^{\pi /4} {\log \left( {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right)d\theta } $

$ = \int_0^{\pi /4} {\log \left( {{2 \over {1 + \tan \theta }}} \right)d\theta = \log 2\int_0^{\pi /4} {1d\theta - I} } $

$ \Rightarrow 2I = {\pi \over 4}\log 2 \Rightarrow I = {\pi \over 8}\log 2$

$ \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} $

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

$ 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} $

We have,

$I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {\tan x + \cot x + \cos ec\,x + \sec x}}} $

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

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{\sin x\cos xdx} \over {1 + \sin x + \cos x}}} $

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

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

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

$ \Rightarrow I = {1 \over 2}[ - \cos x + \sin x - x]_0^{\pi /2}$

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

$ \Rightarrow I = {1 \over 2}\left[ {0 + 1 - {\pi \over 2} + 1} \right] = 1 - {\pi \over 4}$