Question

If $I=\int \frac{dx}{{\sin }^{2}x{\cos }^{2}x}$, then $I$ equals

• A. $-\frac{1}{2}\cot 2x+C$
• B. $-2\cot 2x+C$
• C. $\tan x+\cot x+C$
• D. $\cot x-\tan x+C$

Answer 1

The correct option is B $-2\cot 2x+C$

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

$=4\int \frac{dx}{{\left(2\sin x\cos x\right)}^2}$

$=4\int \frac{dx}{{\left(\sin 2x\right)}^2}$

$=4\int co{\sec }^{2}2xdx$

$=-2\cot 2x+c$