Question

Evaluate $\displaystyle \int \frac{\cos 2x}{\sin^2 x \cdot \cos^2 x} dx$

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

Hint:

Use identities to split rational trigonometric functions.

Answer 1

The Correct Option is A

Use $\cos 2x = \cos^2 x - \sin^2 x$: \[ \int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx = \int \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} dx. \] Split: \[ = \int \Big( \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} \Big) dx = \int \csc^2 x\, dx - \int \sec^2 x\, dx. \] So, \[ = -\cot x + \tan x + C = \tan x - \cot x + C. \]