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

Evaluate the indefinite integral
$$\int\frac {\csc^2{x}-2005}{\cos^{2005}{x}} dx$$

I tried multiplying and dividing by $$\sec^2 {x}$$ and then setting $$\tan{x}=y$$ but no good. Then I set $$\cos {x}=t$$ and tried to create $$\sin {x}$$ in the numerator. But the integral which came was also a difficult one.

Thanks!

P.S. I am a high school student so kindly use elementary methods only. Thanks again!

• Can you show the work which you've tried? and specifically where you got stuck when doing it? – Rivasa Aug 1 '14 at 17:58

HINT:

Integrate by parts $$\int(\sec^{2005}x\cdot\csc^2x)dx$$

$$=\sec^{2005}x\int\csc^2x\ dx-\int\left(\frac{d(\sec^{2005}x)}{dx}\int\csc^2x\ dx\right)dx$$

• Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005\int\sec^{2005}{x}dx$ ? – Henry Aug 1 '14 at 18:24
• @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS. – Tunk-Fey Aug 1 '14 at 18:38
• @Samurai, $$\int\left(\frac{d(\sec^{2005}x)}{dx}\int\csc^2x\ dx\right)dx=\int2005\sec^{2004}x(\sec x\tan x)(-\cot x)\ dx=-2005\int\sec^{2005}x\ dx,$$ right? – lab bhattacharjee Aug 2 '14 at 4:06
• @labbhattacharjee Oh right! Thanks a lot! – Henry Aug 2 '14 at 20:26

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

$$\int \frac{d}{dx}\bigg(\frac{-\cot x}{\cos^{2005}x}\bigg)dx = -\frac{\cot x}{\cos^{2005}x}+C$$