$$\begin{aligned} \text{Let}\quad I & =\int \frac{d x}{1+\cos x}=\int \frac{d x}{1+2 \cos ^2 \frac{x}{2}-1} \\ & =\frac{1}{2} \int \frac{1}{\cos ^2 \frac{x}{2}} d x=\frac{1}{2} \int \sec ^2 \frac{x}{2} d x \\ & =\frac{1}{2} \cdot \tan \frac{x}{2} \cdot 2+C=\tan \frac{x}{2}+C \quad\left[\because \int \sec ^2 x d x=\tan x\right] \end{aligned}$$

Let $$I=\int \tan ^2 x \sec ^4 x d x$$

$$\begin{array}{lrl} \text { Put } & \tan x & =t \Rightarrow \sec ^2 x d x=d t \\ \therefore & I & =\int t^2\left(1+t^2\right) d t=\int\left(t^2+t^4\right) d t \end{array}$$

$=\frac{t^3}{3}+\frac{t^5}{5}+C=\frac{\tan ^5 x}{5}+\frac{\tan ^3 x}{3}+C$

$$\begin{aligned} \text { Let } \quad I & =\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x=\int \frac{(\sin x+\cos x)}{\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}} d x \\ & =\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^2}} d x=\int 1 d x=x+C \end{aligned} $$

$$\begin{aligned} \text{Let}\quad I & =\int \sqrt{1+\sin x} d x \\ & =\int \sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \quad\left[\because \sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}=1\right] \\ & =\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2} d x=\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ & =-\cos \frac{x}{2} \cdot 2+\sin \frac{x}{2} \cdot 2+C=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C \end{aligned}$$

$$\begin{aligned} \text { Let }\quad &I=\int \frac{x}{\sqrt{x}+1} d x\\ &\text { Put } \quad \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & d x & =2 \sqrt{x} d t \\ \therefore & I & =2 \int\left(\frac{x \sqrt{x}}{t+1}\right) d t=2 \int \frac{t^2 \cdot t}{t+1} d t=2 \int \frac{t^3}{t+1} d t \end{array}$$

$$\begin{aligned} & =2 \int \frac{t^3+1-1}{t+1} d t=2 \int \frac{(t+1)\left(t^2-t+1\right)}{t+1} d t-2 \int \frac{1}{t+1} d t \\ & =2 \int\left(t^2-t+1\right) d t-2 \int \frac{1}{t+1} d t \\ & =2\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\log |(t+1)|\right]+C \\ & =2\left[\frac{x \sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |(\sqrt{x}+1)|\right]+C \end{aligned}$$