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) $\int(1-\cos x) cosec^2 x d x=\int cosec^2 x\, d x-\int \cos x\,cosec^2x\,dx$

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

$=\int\frac{1}{\sin^2x}dx-\int\frac{\cos x}{\sin^2x}dx$

$=\tan x-\cot x+C$ (II)