Question

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

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

Answer 1

Answer: (B)

$\int \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \left(\frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}\right)dx$
$=\int \left({\sec }^{2}x-{\text{cosec}}^{2}x\right)$
$=\tan x+\cot x+C$