Question

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

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

Answer 1

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

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

$=\int ({\sec }^{2}x-co{\sec }^{2}x)dx$

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