Question

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

• A. $\tan x-\cot x+c$
• B. $\tan x-x+1$
• C. $\tan x-x$
• D. $\tan x+x$

Answer 1

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

$\int \frac{4dx}{4{\sin }^{2}x{\cos }^{2}x}$

$=\int \frac{4dx}{{\sin }^{2}2x}$

$=4\int {\text{cosec}}^{2}2xdx$

$=-2\cot 2x+C$

$=\frac{-2\left(\cos 2x\right)}{\sin 2x}+C$

$=\frac{2({\sin }^{2}x-{\cos }^{2}x)}{2\sin x.\cos x}+C$

$=\frac{{\sin }^{2}x}{\sin x.\cos x}-\frac{{\cos }^{2}x}{\sin x.\cos x}+C$

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