Question

$\int \left(\frac{{\text{cosec}}^{2}x-2005}{{\cos }^{2005}x}\right)dx$ is equal to
$($where $C$ is the constant of integration$)$

• A. $\frac{\tan x}{{\cos }^{2005}x}+C$
• B. $\frac{\cot x}{{\cos }^{2006}x}+C$
• C. $\frac{-\tan x}{{\cos }^{2005}x}+C$
• D. $-\frac{\cot x}{{\cos }^{2005}x}+C$

Answer 1

The correct option is D $-\frac{\cot x}{{\cos }^{2005}x}+C$

Given, $I=\int \left(\frac{{\text{cosec}}^{2}x-2005}{{\cos }^{2005}x}\right)dx$
$=\int {\text{cosec}}^{2}x\cdot {\cos }^{-2005}xdx-2005\int {\cos }^{-2005}xdx$
Applying by-parts
$={\cos }^{-2005}x(-\cot x)+2005\int {\cos }^{-2006}x(-\sin x)(-\cot x)dx-2005\int {\cos }^{-2005}dx$
$=\frac{-\cot x}{{\cos }^{2005}x}+C$