Question

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

Options

• tan x + cot x + C

• tan x + cosec x + C

• − tan x + cot x + C

• tan x + sec x + C

Answer 1

${\displaystyle \int \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx}$ is equal to tan x + cot x + C.

Explanation:

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

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

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

${\displaystyle =\int ({\sec }^{2}x-\cos e{c}^{2}x) dx}$

${\displaystyle =\tan x+\cot x+C}$