Question

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

• A. $\frac{-1}{\tan x}+\tan x+c$
• B. $\frac{-1}{\tan x}-\tan x+c$
• C. $\frac{-1}{\cot x}+\tan x+c$
• D. None of these

Answer 1

The correct option is A $\frac{-1}{\tan x}+\tan x+c$

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

$={\int }^{}\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$ (since ${\sin }^{2}x+{\cos }^{2}x=1)$

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

$={\int }^{}\left({\sec }^{2}x+\cos e{c}^{2}x\right)dx$

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

$=\tan x-\frac{1}{\tan x}+c$