Question

Solve: $\int \frac{\cos 2x}{{\sin }^{2}x.{\cos }^{2}x}dx$

• A. $\cot x+\tan x+c$
• B. $-\cot x-\tan x+c$
• C. $\cot x-\tan x+c$
• D. $-\cot x+\tan x+c$

Answer 1

The correct option is B $-\cot x-\tan x+c$

$I=\int \frac{\cos 2x}{{\sin }^{2}x.{\cos }^{2}x}dx$

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

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

$=-\cot x-\tan x+c$