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

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

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

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

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

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

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

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

Let $\quad I=\int \sqrt{\frac{a+x}{a-x}} d x$

Put $$x=\operatorname{acos} 2 \theta$$

$$\begin{array}{ll} \Rightarrow & d x=-a \cdot \sin 2 \theta \cdot 2 \cdot d \theta \\ \therefore & I=-2 \int \sqrt{\frac{a+a \cos 2 \theta}{a-a \cos 2 \theta}} \cdot a \sin 2 \theta d \theta \end{array}$$

$\left[\because \cos 2 \theta=\frac{x}{a} \Rightarrow 2 \theta=\cos ^{-1} \frac{x}{a} \Rightarrow \theta=\frac{1}{2} \cos ^{-1} \frac{x}{a}\right]$

$$\begin{aligned} & =-2 a \int \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}} \sin 2 \theta d \theta=-2 a \int \sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}} \sin 2 \theta d \theta \\ & =-2 a \int \cot \theta \cdot \sin 2 \theta d \theta=-2 a \int \frac{\cos \theta}{\sin \theta} \cdot 2 \sin \theta \cdot \cos \theta d \theta \\ & =-4 a \int \cos ^2 \theta d \theta=-2 a \int(1+\cos 2 \theta) d \theta \\ & =-2 a\left[\theta+\frac{\sin 2 \theta}{2}\right]+C \\ & =-2 a\left[\frac{1}{2} \cos ^{-1} \frac{x}{a}+\frac{1}{2} \sqrt{1-\frac{x^2}{a^2}}\right]+C \\ & =-a\left[\cos ^{-1}\left(\frac{x}{a}\right)+\sqrt{1-\frac{x^2}{a^2}}\right]+C \end{aligned}$$

Alternate Method

Let $$I=\int \sqrt{\frac{a+x}{a-x}} d x=\int \sqrt{\frac{(a+x)(a+x)}{(a-x)(a+x)}} d x$$

$$\begin{aligned} & =\int \frac{(a+x)}{\sqrt{a^2-x^2}} d x \\ I & =\int \frac{a}{\sqrt{a^2-x^2}}+\int \frac{x}{\sqrt{a^2-x^2}} d x \end{aligned}$$

$\therefore \quad I=I_1+I_2\quad\text{..... (i)}$

Now, $$I_1=\int \frac{a}{\sqrt{a^2-x^2}}=a \sin ^{-1}\left(\frac{x}{a}\right)+C_1$$

and $$I_2=\int \frac{x}{\sqrt{a^2-x^2}} d x$$

Put $\quad a^2-x^2=t^2 \Rightarrow-2 x d x=2 t d t$

$\therefore \quad I_2=-\int \frac{t}{t} d t e=-\int 1 d t$

$$\begin{array}{rlrl} & =-t+C_2=-\sqrt{a^2-x^2}+C_2 & \\ \therefore\quad I & =a \sin ^{-1}\left(\frac{x}{a}\right)+C_1-\sqrt{a^2-x^2}+C_2 & & {\left[\because t^2=a^2-x^2\right]} \\ I & =a^{\sin ^{-1}}\left(\frac{x}{a}\right)-\sqrt{a^2-x^2}+C & & {\left[\because C=C_1+C_2\right]} \end{array}$$