Question

$\int \frac{x^2+{\cos }^{2}x}{x^2+1}{\text{cosec}}^{2}xdx$ is equal to

• A. $\cot x-{\cot }^{-1}x+c$
• B. $c-\cot x+{\cot }^{-1}x$
• C. ${\tan }^{-1}x-\frac{\text{cosec}x}{\sec x}+c$
• D. $-e^{\log {\tan }^{-1}x}-\cot x+c$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $c-\cot x+{\cot }^{-1}x$
C ${\tan }^{-1}x-\frac{\text{cosec}x}{\sec x}+c$
D $-e^{\log {\tan }^{-1}x}-\cot x+c$

Let
$I=\int \frac{x^2+{\cos }^{2}x}{x^2+1}{cosec}^{2}xdx$
$=\int \frac{x^2+1-1+{\cos }^{2}x}{x^2+1}{cosec}^{2}xdx$
$=\int (1-\frac{{\sin }^{2}x}{x^2+1}){cosec}^{2}xdx$
$=\int ({cosec}^{2}x-\frac{1}{x^2+1})dx$
$=-\cot x-{\tan }^{-1}x+c\cdots \left(1\right)$
$=-\cot x+{\cot }^{-1}x-\frac{\pi }{2}+c$
$=-\cot x+{\cot }^{-1}x+c$

From (1),
$I=-{\tan }^{-1}x-\frac{cosecx}{\sec x}+c$

Again from (1)
$I=-e^{\log {\tan }^{-1}x}-\cot x+c$