$$\begin{aligned} \text{Let}\quad I & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x \\ & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^2 \sqrt{1+\cos x}} d x \\ & =\int_{\pi 3}^{\pi 2} \frac{1}{\left(1-\cos ^2 x\right)} d x=\int_{\pi 3}^{\pi 2} \frac{1}{\sin ^2 x} d x \\ & =\int_{\pi 3}^{\pi 2} \operatorname{cosec}^2 x d x=[-\cot x]_{\pi / 3}^{\pi / 2} \\ & =-\left[\cot \frac{\pi}{2}-\cot \frac{\pi}{3}\right]=-\left[0-\frac{1}{\sqrt{3}}\right]=+\frac{1}{\sqrt{3}} \end{aligned}$$

Alternate Method

Let $I=\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x=\int_{\pi / 3}^{\pi / 2} \frac{\left(2 \cos ^2 \frac{x}{2}\right)^{1 / 2}}{\left(2 \sin ^2 \frac{x}{2}\right)^{5 / 2}} d x$

$=\frac{\sqrt{2}}{4 \sqrt{2}} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x=\frac{1}{4} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x$

$$\begin{array}{lr} \text { Put } & \sin \frac{x}{2}=t \\ \Rightarrow & \cos \frac{x}{2} \cdot \frac{1}{2} d x=d t \\ \Rightarrow & \cos \frac{x}{2} d x=2 d t \end{array}$$

As $\quad x \rightarrow \frac{\pi}{3}$, then $t \rightarrow \frac{1}{2}$

and $x \rightarrow \frac{\pi}{2}$, then $t \rightarrow \frac{1}{\sqrt{2}}$

$$\begin{aligned} \therefore\quad I & =\frac{2}{4} \int_{1 / 2}^{1 \sqrt{2}} \frac{d t}{t^5}=\frac{1}{2}\left[\frac{t^{-5+1}}{-5+1}\right]_{1 / 2}^{1 / \sqrt{2}} \\ & =-\frac{1}{8}\left[\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4}-\frac{1}{\left(\frac{1}{2}\right)^4}\right]^8 \\ & =-\frac{1}{8}(4-16)=\frac{12}{8}=\frac{3}{2} \end{aligned}$$

Let $$I = \int {\mathop {{e^{ - 3x}}}\limits_{II} {{\cos }^3}x\,\mathop {dx}\limits_I } $$

$$\begin{aligned} & =\cos ^3 x \int e^{-3 x} d x-\int\left(\frac{d}{d x} \cos ^3 x \int e^{-3 x} d x\right) d x \\ & =\cos ^3 x \cdot \frac{e^{-3 x}}{-3}-\int\left(-3 \cos ^2 x\right) \sin x \cdot \frac{e^{-3 x}}{-3} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \cos ^2 x \sin x e^{-3 x} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int\left(1-\sin ^2 x\right) \sin x e^{-3 x} d x \end{aligned}$$

$=-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x+$ $$\int {\mathop {{{\sin }^3}}\limits_I x\,{e^{ - 3x}}\mathop {dx}\limits_{II} } $$

$$\begin{aligned} & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x+\sin ^3 x \cdot \frac{e^{-3 x}}{-3}-\int 3 \sin ^2 x \cos x \cdot \frac{e^{-3 x}}{-3} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x-\frac{1}{3} \sin ^3 x e^{-3 x}+\int\left(1-\cos ^2 x\right) \cos x e^{-3 x} d x \\ I & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x}-\frac{1}{3} \sin ^3 x e^{-3 x}+\int \cos x e^{-3 x} d x-\int \cos ^3 x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{3}\left[\cos ^3 x+\sin ^3 x\right]-\left[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x\right]+\int \cos x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{-3}\left[\cos ^3 x+\sin ^3 x\right]+\frac{1}{3} \sin x \cdot e^{-3 x}-\frac{1}{3} \int \cos x \cdot e^{-3 x} d x+\int \cos x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{-3}\left[\cos ^3 x+\sin ^3 x\right]+\frac{1}{3} \sin x e^{-3 x}+\frac{2}{3} \int \cos x e^{-3 x} d x \end{aligned}$$

Now, Let $${I_1} = \int {\mathop {\cos }\limits_I x\,{e^{ - 3x}}\mathop {dx}\limits_{II} } $$

$$\begin{aligned} & I_1=\cos x \cdot \frac{e^{-3 x}}{-3}-\int(-\sin x) \cdot \frac{e^{-3 x}}{-3} d x \\ & I_1=\frac{-1}{3} \cos x \cdot e^{-3 x}-\frac{1}{3} \int \sin x \cdot e^{-3 x} d x \end{aligned}$$

$$\begin{aligned} & =-\frac{1}{3} \cos x \cdot e^{-3 x}-\frac{1}{3}\left[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x\right] \\ & =-\frac{1}{3} \cos x \cdot e^{-3 x}+\frac{1}{9} \sin x \cdot e^{-3 x}-\frac{1}{9} \int \cos x \cdot e^{-3 x} d x \end{aligned}$$

$$\begin{aligned} I_1+\frac{1}{9} I_1 & =-\frac{1}{3} e^{-3 x} \cdot \cos x+\frac{1}{9} \sin x \cdot e^{-3 x} \\ \left(\frac{10}{9}\right) I_1 & =-\frac{1}{3} e^{-3 x} \cdot \cos x+\frac{1}{9} \sin x \cdot e^{-3 x} \\ I_1 & =\frac{-3}{10} e^{-3 x} \cdot \cos x+\frac{1}{10} e^{-3 x} \sin x \end{aligned}$$

$\begin{aligned} 2 I=-\frac{1}{3} e^{-3 x}\left[\sin ^3 x+\cos ^3 x\right]+\frac{1}{3} \sin x \cdot e^{-3 x}-\frac{3}{10} e^{-3 x} & \cdot \cos x \\ & +\frac{1}{10} e^{-3 x} \cdot \sin x+C\end{aligned}$

$$\begin{aligned} \therefore\quad & I=-\frac{1}{6} e^{-3 x}\left[\sin ^3 x+\cos ^3 x\right]+\frac{13}{30} e^{-3 x} \cdot \sin x-\frac{3}{10} e^{-3 x} \cdot \cos x+C \\ & \qquad\left[\begin{array}{l} \because \sin 3 x=3 \sin x-4 \sin ^3 x \\ \text { and } \cos 3 x=4 \cos ^3 x-3 \cos x \end{array}\right] \\ & =\frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x]+\frac{3 e^{-3 x}}{40}[\sin x-3 \cos x]+C \end{aligned}$$

Let $$I=\int \sqrt{\tan x} d x$$

Put $\tan x=t^2 \Rightarrow \sec ^2 x d x=2 t d t$

$$\begin{aligned} \therefore\quad I & =\int t \cdot \frac{2 t}{\sec ^2 x} d t=2 \int \frac{t^2}{1+t^4} d t \\ & =\int \frac{\left(t^2+1\right)+\left(t^2-1\right)}{\left(1+t^4\right)} d t \\ & =\int \frac{t^2+1}{1+t^4} d t+\int \frac{t^2-1}{1+t^4} d t \\ & =\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t+\int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t \\ & =\int \frac{1-\left(-\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t}\right)^2+2}+\int \frac{1+\left(-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}\right)^2-2} d t \end{aligned}$$

Put $$u=t-\frac{1}{t} \Rightarrow d u=\left(1+\frac{1}{t^2}\right) d t$$

and $$v=t+\frac{1}{t} \Rightarrow d v=\left(1-\frac{1}{t^2}\right) d t$$

$$\begin{aligned} \therefore\quad I & =\int \frac{d u}{u^2+(\sqrt{2})^2}+\int \frac{d v}{v^2-(\sqrt{2})^2} \\ & =\frac{1}{\sqrt{2}} \tan ^{-1} \frac{u}{\sqrt{2}}+\frac{1}{2 \sqrt{2}} \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C \\ & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\frac{1}{2 \sqrt{2}} \log \left|\frac{\tan x-\sqrt{2 \tan x}+1}{\tan x+\sqrt{2 \tan x}+1}\right|+C \end{aligned}$$

Let $I=\int_0^{\pi / 2} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}$

$$\begin{aligned} &\text { Divide numerator and denominator by } \cos ^4 x \text {, we get }\\ &\begin{aligned} I & =\int_0^{\pi / 2} \frac{\sec ^4 x d x}{\left(a^2+b^2 \tan ^2 x\right)^2} \\ & =\int_0^{\pi / 2} \frac{\left(1+\tan ^2 x\right) \sec ^2 x d x}{\left(a^2+b^2 \tan ^2 x\right)^2} \end{aligned} \end{aligned}$$

Put $$\tan x=t$$

$\Rightarrow \quad \sec ^2 x d x=d t$

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

and $x \rightarrow \frac{\pi}{2}$, then $t \rightarrow \infty$

$I=\int_0^{\infty} \frac{\left(1+t^2\right)}{\left(a^2+b^2 t^2\right)^2}$

Now, $\frac{1+t^2}{\left(a^2+b^2 t^2\right)^2} \quad\left[\right.$ let $\left.t^2=u\right]$

$$\begin{aligned} &\Rightarrow \quad \frac{1+u}{\left(a^2+b^2 u\right)^2} =\frac{A}{\left(a^2+b^2 u\right)}+\frac{B}{\left(a^2+b^2 u\right)^2} \\ & \Rightarrow \quad 1+u =A\left(a^2+b^2 u\right)+B \end{aligned}$$

On comparing the coefficient of x and constant term on both sides, we get

$$\begin{aligned} a^2 A+B & =1 \quad\text{.... (i)}\\ \text{and}\quad b^2 A & =1 \quad\text{.... (ii)}\\ \therefore\quad A & =\frac{1}{b^2} \end{aligned}$$

Now, $\frac{a^2}{b^2}+B=1$

$\Rightarrow \quad B=1-\frac{a^2}{b^2}=\frac{b^2-a^2}{b^2}$

$\therefore\quad I=\int_0^{\infty} \frac{\left(1+t^2\right)}{\left(a^2+b^2 t^2\right)^2}$

$=\frac{1}{b^2} \int_0^{\infty} \frac{d t}{a^2+b^2 t^2}+\frac{b^2-a^2}{b^2} \int_0^{\infty} \frac{d t}{\left(a^2+b^2 t^2\right)^2}$

$$\begin{aligned} & =\frac{1}{b^2} \int_0^{\infty} \frac{d t}{b^2\left(\frac{a^2}{b^2}+t^2\right)}+\frac{b^2-a^2}{b^2} \int_0^{\infty} \frac{d t}{\left(a^2+b^2 t^2\right)^2} \\ & =\frac{1}{a b^3}\left[\tan ^{-1}\left(\frac{t b}{a}\right)\right]_0^{\infty}+\frac{b^2-a^2}{b^2}\left(\frac{\pi}{4} \cdot \frac{1}{a^3 b}\right) \\ & =\frac{1}{a b^3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]+\frac{\pi}{4} \cdot \frac{b^2-a^2}{\left(a^3 b^3\right)} \\ & =\frac{\pi}{2 a b^3}+\frac{\pi}{4} \cdot \frac{b^2-a^2}{\left(a^3 b^3\right)} \\ & =\pi\left(\frac{2 a^2+b^2-a^2}{4 a^3 b^3}\right)=\frac{\pi}{4}\left(\frac{a^2+b^2}{a^3 b^3}\right) \end{aligned}$$

Let $$\begin{aligned} I & =\int_0^1 x \log (1+2 x) d x \\ & =\left[\log (1+2 x) \frac{x^2}{2}\right]_0^1-\int \frac{1}{1+2 x} \cdot 2 \cdot \frac{x^2}{2} d x \\ & =\frac{1}{2}\left[x^2 \log (1+2 x)\right]_0^1-\int \frac{x^2}{1+2 x} d x \\ & =\frac{1}{2}[1 \log 3-0]-\left[\int_0^1\left(\frac{x}{2}-\frac{\frac{x}{2}}{1+2 x}\right) d x\right] \\ & =\frac{1}{2} \log 3-\frac{1}{2} \int_0^1 x d x+\frac{1}{2} \int_0^1 \frac{x}{1+2 x} d x \\ & =\frac{1}{2} \log 3-\frac{1}{2}\left[\frac{x^2}{2}\right]_0^1+\frac{1}{2} \int_0^1 \frac{1}{2}(2 x+1-1) \\ & =\frac{1}{2} \log 3-\frac{1}{2}\left[\frac{1}{2}-0\right]+\frac{1}{4} \int_0^1 d x-\frac{1}{4} \int_0^1 \frac{1}{1+2 x} d x \\ & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}[x]_0^1-\frac{1}{8}[\log |(1+2 x)|]_0^1 \\ & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}[\log 3-\log 1] \\ & =\frac{1}{2} \log 3-\frac{1}{8} \log 3 \\ & =\frac{3}{8} \log 3 \end{aligned}$$