Question

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

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

Answer 1

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

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

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

$=\int ({\sec }^{2}x+{\csc }^{2}x)dx$

$=\int {\sec }^{2}xdx+\int {\csc }^{2}xdx$

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

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

Hence, the answer is $\tan x-\frac{1}{\tan x}+c.$