Question

Evaluate $\int \frac{1-{\cot }^{2}x}{1+{\cot }^{2}x}dx$
(where $C$ is constant of integration)

• A. $-\frac{\sin 2x}{2}+C$
• B. $\frac{\sin 2x}{2}+C$
• C. $\frac{\cos 2x}{2}+C$
• D. $-\frac{\cos 2x}{2}+C$

Answer 1

The correct option is A $-\frac{\sin 2x}{2}+C$

Let $I=\int \frac{1-{\cot }^{2}x}{1+{\cot }^{2}x}dx$
$=\int \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

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