Question

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

is equal to

• A. $\tan x+\sec x+C$
• B. $\tan x+\cot x+C$
• C. $\tan x+\text{cosec x}+C$
• D. $\tan x+\cot x+C$

Answer 1

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

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

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

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

$=\int {\sec }^{2}x.dx-\int {\text{cosec}}^{2}x.dx$

$=\tan x-(-\cot x)+C$

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

Where $C$ is constant of integration.
Hence, Correct answer is $\left(A\right)$