Question

The value of $\int \frac{cose{c}^{2}x-2005}{{\cos }^{2005}x}dx=$

• A. $\frac{-\cot x}{(\cos x{)}^{2005}}+C$
• B. $\frac{\tan x}{(cosx{)}^{2005}}+C$
• C. $\frac{-\tan x}{(\cos x{)}^{2005}}+C$
• D. $(-\cot x{)}^{2007}+C$

Answer 1

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

$\int \frac{{\csc }^{2}x-2005}{{\cos }^{2005}x}dx$

$\int {\csc }^{2}x{\sec }^{2005}xdx-\int 2005{\sec }^{2004}x\sec x\tan x(-\cot x)dx-\int 2005{\sec }^{2005}xdx+c$

$=-\cot x{\sec }^{2005}x+2005\int {\sec }^{2005}xdx-2005\int {\sec }^{2005}xdx+c$

$=-\cot x{\sec }^{2005}x+c$

[ using $\int uvdx=u\int vdx-\int \left(\frac{d}{dx}\right(u\left)\int vdx\right)dx$

where $u={\sec }^{2005}x$

$\theta ={\csc }^{2}x$