Column - I $f\left(x\right)$		Column - II $\int f\left(x\right)dx$
a.	$\frac{1}{\left(a^2+b^2\right)-\left(a^2-b^2\right)\cos x}$	p.	$\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan \frac{x}{2}\right)+c$
b.	$\frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}$	q.	$\frac{1}{a^2\sin \alpha }{\tan }^{-1}\left(\frac{\tan x}{\sin \alpha }\right)+c,\alpha ={\cos }^{-1}\frac{b}{a}$
c.	$\frac{1}{a\cos x+b\sin x}$	r.	$\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan x\right)+c$
d.	$\frac{1}{a^2-b^2{\cos }^{2}x}$	s.	$\frac{1}{\sqrt{a^2+b^2}}\log \left|\tan \frac{1}{2}\left(x+{\tan }^{-1}\frac{a}{b}\right)\right|+c$

(1)
 $$\begin{gathered} I=\int \frac{1}{\left(a^2+b^2\right)-\left(a^2-b^2\right)\cos x}dx \\ put \tan \frac{x}{2}=t\Rightarrow dx=\frac{2dt}{1+t^2},\cos x=\frac{1-t^2}{1+t^2} \end{gathered}$$

$$\begin{gathered} =2\int \frac{dt}{\left(a^2+b^2\right)\left(1+t^2\right)-\left(a^2-b^2\right)\left(1-t^2\right)} \\ =2\int \frac{dt}{\left(a^2\right)\left(2t^2\right)+\left(b^2\right)\left(2\right)} \\ =\int \frac{dt}{{\left(at\right)}^2+b^2}=\frac{1}{ab}{\tan }^{-1}\frac{at}{b}=\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan \frac{x}{2}\right)+c \end{gathered}$$

(2)
 $$\begin{gathered} \int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx=\int \frac{{\sec }^{2}x}{b^2+a^2{\tan }^{2}x}dx \\ =\int \frac{dt}{b^2+a^2{t}^2} put \tan x=t \\ =\frac{1}{ab}{\tan }^{-1}\frac{at}{b}=\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan x\right)+c \end{gathered}$$

(3)
 $$\begin{gathered} I=\int \frac{1}{a\cos x+b\sin x}dx \\ =\frac{1}{\sqrt{a^2+b^2}}\int \frac{1}{{\displaystyle \frac{a}{\sqrt{a^2+b^2}}}\cos x+{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}\sin x}dx \\ =\frac{1}{\sqrt{a^2+b^2}}\int \frac{1}{{\displaystyle \sin \theta \cos x+\cos \theta \sin x}}dx \\ where \frac{a}{\sqrt{a^2+b^2}}=\sin \theta ,\frac{b}{\sqrt{a^2+b^2}}=\cos \theta \\ I=\frac{1}{\sqrt{a^2+b^2}}\int \mathrm{cosec}(\theta +x)dx \\ =\frac{1}{\sqrt{a^2+b^2}}\log \left|\tan \frac{(\theta +x)}{2}\right| \\ =\frac{1}{\sqrt{a^2+b^2}}\log \left|\tan \frac{1}{2}\left(x+{\tan }^{-1}\frac{a}{b}\right)\right|+c \end{gathered}$$

(4)
 $$\begin{gathered} \int \frac{{\sec }^{2}x}{a^2\left(1+{\tan }^{2}x\right)-b^2}dx\text{ put }\tan x=t \\ =\int \frac{dt}{a^2\left(1+t^2\right)-b^2} \\ =\int \frac{dt}{a^2\left(t^2\right)+a^2-b^2} \\ =\frac{1}{a^2\sin \alpha }{\tan }^{-1}\left(\frac{\tan x}{\sin \alpha }\right)+c,\alpha ={\cos }^{-1}\frac{b}{a} \end{gathered}$$