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$$I=\int_0^{\frac{\pi}{2}} \frac{\cos^{2008} (x)}{\sin^{2008} (x)+\cos^{2008} (x)}~dx$$

I hope anyone can answer this definite integral.

projectilemotion
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user420237
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    This was most likely a contest problem from 2008 lol – Saketh Malyala Feb 26 '17 at 17:57
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    As a general rule, when you see a number like 2008 in a question, it's a good clue that the number isn't important. For cosines and sines, all that usually matters is whether the exponent is odd or even - so try replacing all of the 2008's with 2's, and see if you can solve it. Then try it with 4 or 6, and see if you see a pattern you can use for 2008. – Reese Johnston Feb 26 '17 at 17:58
  • It was 1/(1+tan^2008 x). I managed to make it in this way.(i donot know any more) – user420237 Feb 26 '17 at 17:58
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    Same way as: http://math.stackexchange.com/questions/1305511/solve-trigonometric-integral-int-pi-2-pi-2-frac-sin2014x-sin20 – wythagoras Feb 26 '17 at 18:00
  • $\dfrac{\pi}{2}-x=u$ – Nosrati Feb 26 '17 at 18:00
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    Alternatively, http://math.stackexchange.com/questions/605673/integrate-int-0-pi-2-frac11-tan-alphax-mathrmdx?noredirect=1&lq=1 – wythagoras Feb 26 '17 at 18:01
  • Haha, 2008 is actually a hint, since this number is insanely large for any serious manipulation. – Violapterin Feb 26 '17 at 18:16

2 Answers2

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Enforcing the substitution $x\to \pi/2-x$ yields

$$\int_0^{\pi/2}\frac{\cos^{2008}(x)}{\cos^{2008}(x)+\sin^{2008}(x)}\,dx=\int_0^{\pi/2}\frac{\sin^{2008}(x)}{\cos^{2008}(x)+\sin^{2008}(x)}\,dx$$

Hence,

$$\int_0^{\pi/2}\frac{\cos^{2008}(x)}{\cos^{2008}(x)+\sin^{2008}(x)}\,dx=\frac12\int_0^{\pi/2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\cos^{2008}(x)+\sin^{2008}(x)}\,dx=\pi/4$$

Mark Viola
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2

Hint...substitute $u=\frac{\pi}{2}-x$ and you get the same integral but with $\sin$ in the numerator. So $2I=$?

David Quinn
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