Question

Integrate:

$\int \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$ is equal to ?

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

Answer 1

The correct option is A $\tan x+\cot x+C$

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

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

$=\int {\sec }^{2}x.dx-\int cose{c}^{2}x.dx$

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