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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.

Please Help!

Thanks!

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

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  • $\begingroup$ Can you show the work which you've tried? and specifically where you got stuck when doing it? $\endgroup$
    – Rivasa
    Aug 1, 2014 at 17:58

2 Answers 2

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HINT:

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

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

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    $\begingroup$ Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005\int\sec^{2005}{x}dx$ ? $\endgroup$
    – Henry
    Aug 1, 2014 at 18:24
  • $\begingroup$ @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS. $\endgroup$
    – Tunk-Fey
    Aug 1, 2014 at 18:38
  • $\begingroup$ @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? $\endgroup$ Aug 2, 2014 at 4:06
  • $\begingroup$ @labbhattacharjee Oh right! Thanks a lot! $\endgroup$
    – Henry
    Aug 2, 2014 at 20:26
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$$\int \frac{\csc^2 x-2005}{\cos^{2005}x}dx = \int \frac{\cos^{2005}x\csc^2 x-2005\cos^{2005}x}{(\cos^{2005}x)^2}dx$$

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

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