Match List – I with List – II.
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LIST I |
LIST II |
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A. $\int \sin x \cos x d x$ |
I. $\log _e\left(1+\sin ^2 x\right)+C$ |
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B. $\int \frac{d x}{\sin ^2 \cos ^2 x}$ |
II. $\tan \frac{x}{2}+C$ |
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C. $\int \frac{\sin 2x}{1+\sin ^2 x} d x$ |
III. $-\frac{1}{4} \cos 2 x+C$ |
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D. $\int(1-\cos x) cosec^2 x d x$ |
IV. $\tan x-\cot x+C$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
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)