Definite Integration

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x \\ & f: R \rightarrow R, f(x)=x^2+2 \\ & \text { Now, } \lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right) \\ +\ldots+f(7) \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} \left(\frac{7}{n}\right)^2+2+\left(\frac{14}{n}\right)^2+2+\left(\frac{21}{n}\right)^2 \\ +2+\ldots+\left(\frac{7 n}{n}\right)^2+2 \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} 2 n+\left(\frac{7}{n}\right)^2\left(1^2+2^2+3^2\right. \\ +\ldots+n^2 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[2 n+\frac{49}{n^2} \frac{n(n+1)(2 n+1)}{6}\right] \\ & =\lim _{n \rightarrow \infty} 3\left[2+\frac{49}{6} \frac{n^2\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{n^2}\right] \\ & =3\left[2+\frac{49}{6} \times 2\right] \\ & =55 \end{aligned} $

Given,

$ \begin{aligned} & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ \sin 2 x & 2 \sin ^2 x & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \\ & \text { Applying } R_2 \rightarrow R_2-2 \cos x R_3 \\ & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ 0 & 2 & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \end{aligned} $

Expanding along $R_2$,

$ \begin{aligned} & f(x)=\left[0+2\left(0-\sin ^2 x\right)\right. \\ &+ \cos x\left(-2 \cos ^3 x-\sin x \sin 2 x\right] \\ & \Rightarrow f(x)=-2 \sin ^2 x \\ &-2 \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right)=-2 \\ & \Rightarrow f^{\prime}(x)=0 \\ & \therefore \int_0^{\pi / 4}\left[2|f(x)|+5 f^{\prime}(x)\right] d x \\ &=\int_0^{\pi / 4}(4+0) d x=\pi \end{aligned} $

$ \text { } \begin{aligned}Let \,\,\, I & =\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x \\ \quad I & =\left[\frac{-\cos \frac{\pi}{3} x}{\frac{\pi}{3}}-\frac{\sin \left(\frac{\pi}{3} x\right)}{\frac{\pi}{3}}\right]_0^3 \end{aligned} $

$ \begin{aligned} & =-\frac{3}{\pi}\left[\cos \frac{\pi}{3} x+\sin \left(\frac{\pi x}{3}\right)\right]_0^3 \\ & =-\frac{3}{\pi}[(\cos \pi+\sin \pi)-(\cos 0+\sin 0)] \\ & =-\frac{3}{\pi}[-1-1]=\frac{6}{\pi} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^{\pi / 2} \sin ^4 \theta \cos ^3 \theta d \theta \\ & =\int_0^{\pi / 2} \sin ^4 \theta+\left(1-\sin ^2 \theta\right) \cos \theta d \theta \end{aligned} $

Let $\sin \theta=t$, then $\cos \theta d \theta=d t$

When $\theta \rightarrow 0, t \rightarrow 0$

$ \begin{aligned} & \text { and } \theta \rightarrow \frac{\pi}{2}, t \rightarrow 1 \\ & \qquad \begin{aligned} I & =\int_0^1 t^4\left(1-t^2\right) d t \\ & =\int_0^1\left(t^4-t^6\right) d t \\ & =\left[\frac{t^5}{5}-\frac{t^7}{7}\right]_0^1=\frac{1}{5}-\frac{1}{7} \\ & =\frac{2}{35} \end{aligned} \end{aligned} $

Let $1+x^2=t$

$2 x d x=d t$

When $x=0$ and $t=1$, when $x=1, t=2$

$ \begin{aligned} \therefore \exp \left[\int_1^2 \log t\right] & =\exp \left[[t \log t-t]_1^2\right] \\ & =\exp [2 \log 2-2+1] \\ & =\exp [\log 4-1] \\ & =e^{\log 4} \times e^{-1}=\frac{4}{e} \end{aligned} $

$ \text { } \begin{aligned}Let\,\, I & =\int_0^{2 a} f(x) d x \\ & =\int_0^a f(x) d x+\int_a^{2 a} f(x) d x \ldots \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Let\,\, I_1 & =\int_a^{2 a} f(x) d x \\ Let\,\,\,x & =t+a \\ d x & =d t \end{aligned} \end{aligned} $

$\,\,\,\,\,\,\,\,\,\, I_1=\int_0^a f(t+a) d t=\int_0^a f(a+x) d x $

$ \therefore \quad I=\int_0^a f(x) d x+\int_0^a f(a+x) d x $

Let $I=\int_1^2 x \sqrt{4-x^2} d x$

On putting $4-x^2=1$,

$ \begin{aligned} & -2 x d x =d t \\ \Rightarrow &\,\,\,\, x d x =-\frac{d t}{2} \end{aligned} $

When $x \rightarrow 1, t \rightarrow 3$

$ \begin{aligned} & x \rightarrow 2, t \rightarrow 0 \\ & I=\frac{-1}{2} \int_3^0 \sqrt{t} d t \\ &=\frac{1}{2}\left[\frac{2}{3}\left(t^{\frac{3}{2}}\right)\right]_0^3 \\ &=\frac{1}{2} \times \frac{2}{3} \times 3 \sqrt{3}=\sqrt{3} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} \text { } k= & \int_{\frac{-3}{2}}^{\frac{3}{2}}[2 x-3] d x=\int_{\frac{-3}{2}}^{\frac{3}{2}}([2 x]-3) d x \\ = & \int_{-3 / 2}^{-1}(-3) d x+\int_{-1}^{-1 / 2}(-2) d x+\int_{\frac{-1}{2}}^0(-1) d x+\int_0^{1 / 2}(0) d x +\int_{\frac{1}{2}}^1 d x+\int_1^{3 / 2} 2 d x-\int_{\frac{-3}{2}}^{3 / 2} 3 d x \\ = & \frac{-3}{2}-1-\frac{1}{2}+\frac{1}{2}+1-9=\frac{-3}{2}-9 \\ \therefore & \left|k+\frac{1}{2}\right|=\left|-\frac{3}{2}-9+\frac{1}{2}\right|=10 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} 1 d x \\ & =\left[\frac{x^2}{2}+\frac{2}{3} x^{\frac{3}{2}}+\ln x\right]_1^4-[x]_1^{2 \log 2} \\ & =\left(8+\frac{16}{3}+\ln 4\right)-\left(\frac{1}{2}+\frac{2}{3}+0\right) \\ & -(2 \log 2-1)=\frac{17}{2}+\frac{14}{3}=\frac{79}{6} \end{aligned} $

Let

$ I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x $

Let

$ \begin{aligned} f(x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \\ f(-x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4-\sin x}{4+\sin x}\right)\right) \\ & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}-\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \end{aligned} $

$ \begin{aligned} f(x) & +f(-x)=\frac{6}{\pi(2-\cos 2 x)} \\ I & =\int_{-\pi / 4}^{\pi / 4} f(x) d x \\ & =\int_0^{\pi / 4} f(x)+f(-x) d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x \\ I & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1+\tan ^2 x}{1+3 \tan ^2 x} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x \end{aligned} $

$ \begin{aligned} &\text { On putting } \tan x=t\\ &\begin{aligned} & \sec ^2 x d x=d t \\ & I=\frac{6}{\pi} \int_0^1 \frac{d t}{1+(\sqrt{3} t)^2}=\frac{6}{\pi} \cdot \frac{1}{\sqrt{3}}\left[\tan ^{-1} \sqrt{3} t\right]_0^1 \\ & =\frac{2 \sqrt{3}}{\pi}\left[\tan ^{-1} \sqrt{3}-0\right] \\ & \quad I=\frac{2 \sqrt{3}}{\pi} \times \frac{\pi}{3}=\frac{2}{\sqrt{3}} \\ & \Rightarrow \quad I^2=\frac{4}{3} \\ & \quad 3 I^2=4 \end{aligned} \end{aligned} $

$ \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} $