^2x.cosec^2xdx=

MCQ

$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}=$

✓
$\tan\text{x}-\cot\text{x+c}$
B
$\tan\text{x}+\cot\text{x+c}$
C
$-\tan\text{x}+\cot\text{x+c}$
D
$\sec\text{x}\tan\text{x+c}$

✓

Answer

Correct option: A.

$\tan\text{x}-\cot\text{x+c}$

$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}$
$=\int\frac{{1}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}}}}}{{\cos^2\text{x}\sin^2\text{x}}}+\frac{{{{\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\sec^2\text{x})\text{dx}$
$=-\cot\text{x}+\tan\text{x}+\text{c}$
$=\tan\text{x}-\cot\text{x+c}$

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