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

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

$$\begin{aligned} I & =\int_0^\pi(\pi-x) \log \sin (\pi-x) d x \\ & =\int_0^\pi(\pi-x) \log \sin x d x \quad\text{.... (ii)}\\ 2 I & =\pi \int_0^\pi \log \sin x d x \quad\text{.... (iii)}\\ 2 I & =2 \pi \int_0^{\pi / 2} \log \sin x d x \quad \left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]\\ I & =\pi \int_0^{\pi / 2} \log \sin x d x \quad\text{.... (iv)}\\ \text{Now,}\quad I & =\pi \int_0^{\pi / 2} \log \sin (\pi / 2-x) d x\quad\text{.... (v)} \end{aligned}$$

On adding Eqs. (iv) and (v), we get

$$\begin{aligned} 2 I & =\pi \int_0^{\pi / 2}(\log \sin x+\log \cos x) d x \\ 2 I & =\pi \int_0^{\pi / 2} \log \sin x \cos x d x \\ & =\pi \int_0^{\pi / 22} \log \frac{2 \sin x \cos x}{2} d x \\ 2 I & =\pi \int_0^{\pi / 2}(\log \sin 2 x-\log 2) d x \\ 2 I & =\pi \int_0^{\pi / 2} \log \sin 2 x d x-\pi \int_0^{\pi / 2} \log 2 d x \end{aligned}$$

Put $2 x=t \Rightarrow d x=\frac{1}{2} d t$

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

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

$$\begin{array}{ll} \therefore & 2 I=\frac{\pi}{2} \int_0^\pi \log \sin t d t-\frac{\pi^2}{2} \log 2 \\ \Rightarrow & 2 I=\frac{\pi}{2} \int_0^\pi \log \sin x d x-\frac{\pi^2}{2} \log 2 \\ \Rightarrow & 2 I=I-\frac{\pi^2}{2} \log 2 \quad\text{[from Eq. (iii)]}\\ \therefore & I=-\frac{\pi^2}{2} \log 2=\frac{\pi^2}{2} \log \left(\frac{1}{2}\right) \end{array}$$

$$\begin{aligned} \text{Let}\quad I & =\int_{-\pi / 4}^{\pi / 4} \log (\sin x+\cos x) d x \quad\text{.... (i)}\\ I & =\int_{-\pi / 4}^{\pi / 4} \log \left\{\sin \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)\right\} d x \\ & =\int_{-\pi / 4}^{\pi / 4} \log \{\sin (-x)+\cos (-x)\} d x \\ \text{and}\quad I & =\int_{-\pi / 4}^{\pi /} \log (\cos x-\sin x) d x\quad\text{.... (ii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned} & 2 I=\int_{-\pi / 4}^{\pi 4} \log \cos 2 x d x \\ & 2 I=\int_0^{\pi / 4} \log \cos 2 x d x\quad\text{.... (iii)} \end{aligned}$$

$\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x)\right.$, if $\left.f(-x)=f(x)\right]$

Put $$2 x=t \Rightarrow d x=\frac{d t}{2}$$

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

and $x \rightarrow \frac{\pi}{4}$, then $t \rightarrow \frac{\pi}{2}$

$$\begin{aligned} & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos t d t \quad\text{.... (iv)}\\ & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos \left(\frac{\pi}{2}-t\right) d t \quad \left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]\\ \Rightarrow\quad & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t d x\quad\text{.... (v)} \end{aligned}$$

On adding Eqs. (iv) and (v), we get

$$\begin{aligned} & 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t \cos t d t \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \frac{\sin 2 t}{2} d t \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin 2 x d x-\frac{1}{2} \int_0^{\pi / 2} \log 2 d x \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin \left(\frac{\pi}{2}-2 x\right) d x-\log 2 \cdot \frac{\pi}{4} \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos 2 x d x-\frac{\pi}{4} \log 2 \end{aligned}$$

$$\begin{array}{llr} \Rightarrow & 4 I=\int_0^{\pi / 4} \log \cos 2 x d x-\frac{\pi}{4} \log 2 & {\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]} \\ \Rightarrow & 4 I=2 I-\frac{\pi}{4} \log 2 & \quad \text { [from Eq. (iii)] } \\ \therefore & I=-\frac{\pi}{8} \log 2=\frac{\pi}{8} \log \left(\frac{1}{2}\right) \end{array}$$