Match List – I with List – II.

Choose the correct answer from the options given below:

A-III, B-IV, C-I, D-II

The correct answer is Option (1) → A-III, B-IV, C-I, D-II

$\int \sin x \cos x d x=\frac{1}{2}\int 2\sin x\cos x\,dx$

$=\frac{1}{2}\sin 2xdx=-\frac{1}{4}\cos 2x+c$ (III)

(B) $\int\frac{dx}{\sin^2x\cos^2x}=\int\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}dx$

$=\int\frac{1}{\cos^2x}dx+\int\frac{1}{\sin^2x}dx$  $[\sin^2+\cos^2x=1]$

$=\tan x-\cot x+c$ (IV)

(C) $\int \frac{\sin  2x}{1+\sin ^2 x} d x=\int\frac{2\sin x\cos x}{1+\sin x}dx$

let $t=\sin x⇒dt=\cos x\,dx$

$I=\int\frac{2t}{1+t^2}dt$

$=\log|1+t^2|+C$

$=\log|1+\sin^2x|+C$ (I)

(D) $\int(1-\cos x) cosec^2 x d x$

$\int \text{cosec}^2 x \, dx = -\cot x$

$\int \frac{\cos x}{\sin^2 x} \, dx = \int \frac{d(\sin x)}{\sin^2 x} = -\frac{1}{\sin x}$

$\int (1 - \cos x) \text{cosec}^2 x \, dx = -\cot x + \text{cosec} x + C$

$\tan \frac{x}{2} = \text{cosec} x - \cot x$

$\tan \frac{x}{2}+C$  (II)