Question

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

• A. $\tan x+\cot x+C$
• B. $\tan x-\cot x+C$
• C. $\tan x\cdot \cot x+C$
• D. $\tan x-\cot 2x+C$

Answer 1

The correct option is B $\tan x-\cot x+C$

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

$=\int \frac{({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x\cdot {\cos }^{2}x}$

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

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

$=\int {\sec }^{2}xdx+\int {\text{cosec}}^{2}xdx$

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