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  • Choose the correct answer integral (dx /(sin^2x cos^2x)

Q39  Choose the correct answer  \int \frac{dx }{\sin ^ 2 x \cos ^2 x }\: \: \: equals 

 (A) \tan x + \cot x + C \\\\ (B) \tan x - \cot x + C\\\\ (C) \tan x \cot x + C\\\\ (D) \tan x - \cot 2x + C

Answers (1)

best_answer

Given integral \int \frac{dx }{\sin ^ 2 x \cos ^ 2x }

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

=\int \frac{\sin ^2 x +\cos^2 x }{\sin^2 x \cos^2 x}dx                                         \left ( \because \sin ^2 x +\cos^2 x =1 \right )

=\int \frac{\sin^2 x }{\sin^2 x \cos^2 x}dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x}dx

=\int \sec^2 x dx + \int cosec^2 x dx

=\tan x -\cot x +C

Therefore, the correct answer is B.

 

Posted by

Divya Prakash Singh

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