Let $$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 4}} d x=\int \frac{\sin ^{-1} x}{\left(1-x^2\right) \sqrt{1-x^2}} d x$$

Put $\quad \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$

and $$x=\sin t \Rightarrow 1-x^2=\cos ^2 t $$

$\Rightarrow \quad \cos t=\sqrt{1-x^2}$

$\therefore \quad I=\int \frac{t}{\cos ^2 t} d t=\int t \cdot \sec ^2 t d t$

$$\begin{aligned} & =t \cdot \int \sec ^2 t d t-\int\left(\frac{d}{d t} t \cdot \int \sec ^2 t d t\right) d t \\ & =t \cdot \tan t-\int 1 \cdot \tan t d t \\ & =t \tan t+\log |\cos t|+C \quad \quad\left[\because \int \tan x d x=-\log |\cos x|+C\right] \\ & =\sin ^{-1} x \cdot \frac{x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|+C \end{aligned}$$

Let $I=\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{1-2\left(2 \cos ^2 \frac{3 x}{2}-1\right)} d x$

$\left[\because \cos C+\cos D=2 \cos \frac{C+D}{2} \cdot \cos \frac{C-D}{2}\right.$ and $\left.\cos 2 x=2 \cos ^2 x-1\right]$

$I=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{3-4 \cos ^2 \frac{3 x}{2}} d x=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{4 \cos ^2 \frac{3 x}{2}-3} d x$

$=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{4 \cos ^3 \frac{3 x}{2}-3 \cos \frac{3 x}{2}} d x \quad\left[\right.$ multiply and divide by $\left.\cos \frac{3 x}{2}\right]$

$$ \begin{aligned} & =-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{\cos 3 \cdot \frac{3 x}{2}} d x=-\int 2 \cos \frac{3 x}{2} \cdot \cos \frac{x}{2} d x \\ & =-\int\left\{\cos \left(\frac{3 x}{2}+\frac{x}{2}\right)+\cos \left(\frac{3 x}{2}-\frac{x}{2}\right)\right\} d x \\ & =-\int(\cos 2 x+\cos x) d x \\ & =-\left[\frac{\sin 2 x}{2}+\sin x\right]+C \\ & =-\frac{1}{2} \sin 2 x-\sin x+C \end{aligned}$$

Let $$\begin{aligned} I & =\int \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} d x=\int \frac{\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3}{\sin ^2 x \cdot \cos ^2 x} d x \\ & =\int \frac{\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x-\sin ^2 x \cos ^2 x+\cos ^4 x\right)}{\sin ^2 x \cdot \cos ^2 x} d x \\ & =\int \frac{\sin ^4 x}{\sin ^2 x \cos ^2 x} d x+\int \frac{\cos ^4 x}{\sin ^2 x \cdot \cos ^2 x} d x-\int \frac{\sin ^2 x \cos ^2 x}{\sin ^2 x \cdot \cos ^2 x} d x \\ & =\int \tan ^2 x d x+\int \cot ^2 x d x-\int 1 d x \\ & =\int\left(\sec ^2 x-1\right) d x+\int\left(\operatorname{cosec}^2 x-1\right) d x-\int 1 d x \\ & =\int \sec ^2 x d x+\int \operatorname{cosec}^2 x d x-3 \int d x \\ I & =\tan x-\cot x-3 x+C \end{aligned}$$

Let $$I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x=\int \frac{\sqrt{x}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(x^{3 / 2}\right)^2}}$$

Put $\quad x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t$

$\therefore \quad I=\frac{2}{3} \int \frac{d t}{\sqrt{\left(a^{3 / 2}\right)^2-t^2}}=\frac{2}{3} \sin ^{-1} \frac{t}{a^{3 / 2}}+C$

$=\frac{2}{3} \sin ^{-1} \frac{x^{3 / 2}}{a^{3 / 2}}+C=\frac{2}{3} \sin ^{-1} \sqrt{\frac{x^3}{a^3}}+C$

Let $$\begin{aligned} I & =\int \frac{\cos x-\cos 2 x}{1-\cos x} d x=\int \frac{2 \sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{1-1+2 \sin ^2 \frac{x}{2}} d x \\ & =2 \int \frac{\sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{2 \sin ^2 \frac{x}{2}} d x=\int \frac{\sin \frac{3 x}{2}}{\sin \frac{x}{2}} d x \end{aligned}$$

$=\int \frac{3 \sin \frac{x}{2}-4 \sin ^3 \frac{x}{2}}{\sin \frac{x}{2}} d x \quad\left[\because \sin 3 x=3 \sin x-4 \sin ^3 x\right]$

$$\begin{aligned} & =3 \int d x-4 \int \sin ^2 \frac{x}{2} d x=3 \int d x-4 \int \frac{1-\cos x}{2} d x \\ & =3 \int d x-2 \int d x+2 \int \cos x d x \\ & =\int d x+2 \int \cos x d x=x+2 \sin x+C=2 \sin x+x+C \end{aligned}$$