Let $I=\int_0^1 \frac{d x}{e^x+e^{-x}}=\int_0^1 \frac{e^x}{1+e^{2 x}} d x$

Put $$e^x=t$$

$\Rightarrow \quad e^x d x=d t$

$\therefore \quad I=\int_1^e \frac{d t}{1+t^2}=\left[\tan ^{-1} t\right]_1^e$

$$\begin{aligned} & =\tan ^{-1} e-\tan ^{-1} 1 \\ & =\tan ^{-1} e-\frac{\pi}{4} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_0^{\pi / 2} \frac{\tan x d x}{1+m^2 \tan ^2 x} d x \\ & =\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{1+m^2 \cdot \frac{\sin ^2 x}{\cos ^2 x}} d x \\ & =\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{\frac{\cos ^2 x+m^2 \sin ^2 x}{\cos ^2 x}} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x d x}{1-\sin ^2 x+m^2 \sin ^2 x} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x}{1-\sin ^2 x\left(1-m^2\right)} d x \end{aligned}$$

$$\begin{array}{lr} \text { Put } & \sin ^2 x=t \\ \Rightarrow & 2 \sin x \cos x d x=d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\frac{1}{2} \int_0^1 \frac{d t}{1-t\left(1-m^2\right)} \\ & =\frac{1}{2}\left[-\log \left|1-t\left(1-m^2\right)\right| \cdot \frac{1}{1-m^2}\right]_0^1 \\ & =\frac{1}{2}\left[-\log \left|1-1+m^2\right| \cdot \frac{1}{1+m^2}+\log |1| \cdot \frac{1}{1-m^2}\right] \\ & =\frac{1}{2}\left[-\log \left|m^2\right| \cdot \frac{1}{1-m^2}\right]=\frac{2}{2} \cdot \frac{\log m}{\left(m^2-1\right)} \\ & =\log \frac{m}{m^2-1} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_1^2 \frac{d x}{\sqrt{(x-1)(2-x)}}=\int_1^2 \frac{d x}{\sqrt{2 x-x^2-2+x}} \\ & =\int_1^2 \frac{d x}{\sqrt{-\left(x^2-3 x+2\right)}} \end{aligned}$$

$$\begin{aligned} & =\int_1^2 \frac{d x}{\sqrt{-\left[x^2-2 \cdot \frac{3}{2} x+\left(\frac{3}{2}\right)^2+2-\frac{9}{4}\right]}} \\ & =\int_1^2 \frac{d x}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2\right\}}} \\ & =\int_1^2 \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}=\left[\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{1}{2}}\right)\right]_1^2 \end{aligned}$$

$$\begin{aligned} & =\left[\sin ^{-1}(2 x-3)\right]_1^2=\sin ^{-1} 1-\sin ^{-1}(-1) \\ & =\frac{\pi}{2}+\frac{\pi}{2} \quad\left[\because \sin \frac{\pi}{2}=1 \text { and } \sin (-\theta)=-\sin \theta\right] \\ & =\pi \end{aligned}$$

Let $I=\int_0^1 \frac{x}{\sqrt{1+x^2}} d x$

Put $1+x^2=t^2$

$$\begin{array}{lr} \Rightarrow & 2 x d x=2 t d t \\ \Rightarrow & x d x=t d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\int_1^{\sqrt{2}} \frac{t d t}{t} \\ & =[t]_1^{\sqrt{2}}=\sqrt{2}-1 \end{aligned}$$

Let $$I=\int_0^\pi x \sin x \cos ^2 x d x\quad\text{.... (i)}$$

and $$I=\int_0^\pi(\pi-x) \sin (\pi-x) \cos ^2(\pi-x) d x$$

$\Rightarrow \quad I=\int_0^\pi(\pi-x) \sin x \cos ^2 x d x\quad\text{.... (ii)}$

On adding Eqs. (i) and (ii), we get

$$2 I=\int_0^\pi \pi \sin x \cos ^2 x d x$$

$$\begin{array}{ll} \text { Put } & \cos x=t \\ \Rightarrow & -\sin x d x=d t \end{array}$$

As $x \rightarrow 0$, then $t \rightarrow 1$

and $x \rightarrow \pi$, then $t \rightarrow-1$

$\therefore\quad I=-\pi \int_1^{-1} t^2 d t \Rightarrow I=-\pi\left[\frac{t^3}{3}\right]_1^{-1}$

$\Rightarrow\quad 2 I=-\frac{\pi}{3}[-1-1] \Rightarrow 2 I=\frac{2 \pi}{3}$

$\therefore \quad I=\frac{\pi}{3}$