Question

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

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

Answer 1

The correct option is A $-\cot x-x+C$

Let $I=\int \frac{1+{\cot }^{2}x}{1+{\tan }^{2}x}dx$
$\Rightarrow I=\int \frac{{\text{cosec}}^{2}x}{{\sec }^{2}x}dx$
$(\because 1+{\cot }^{2}x={\text{cosec}}^{2}x,1+{\tan }^{2}x={\sec }^{2}x)$
$I=\int {\cot }^{2}xdx$

$\Rightarrow I=\int ({\text{cosec}}^{2}x-1)dx=-\cot x-x+C$