Question

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

Answer 1

Consider the given function.

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

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

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

$=\int {\sec }^{2}xdx-\int cose{c}^{2}xdx$

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

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

Hence, this is the answer.