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$$\int_{ - \pi /2}^{\pi /2} \frac1{2007^{x} + 1}\cdot \frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx .$$

This integral stuns me for a while, I just can't solve it! I tried integration by recurrence since here it seems that we have something that looks like a recurrence, but nothing comes up. I also tried all the integration techniques I'm aware of but they just don't work.

So I fear like this integral uses something like Fourier, or some Laplace transform thing or something of that sort but I don't know them.

That's why I will be very thankful if they could explain me how to solve it.

Thank you.

user165134
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1 Answers1

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You can use this way to do. Let $$ I=\int_{ - \pi /2}^{\pi /2} \frac1{2007^{x} + 1}\cdot \frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx.\tag{1}$$ Now use the substitution $x=-u$ to get \begin{eqnarray*} I&=&-\int_{\pi /2}^{-\pi /2} \frac1{2007^{-u} + 1}\cdot \frac {\sin^{2008}(-u)}{\sin^{2008}(-u) + \cos^{2008}(-u)} \, (-du)\\ &=&\int_{-\pi /2}^{\pi /2} \frac{2007^u}{2007^{u} + 1}\cdot \frac {\sin^{2008}u}{\sin^{2008}u + \cos^{2008}u} \, dx.\tag{2} \end{eqnarray*} So adding (1) and (2) will give $$ 2I=\int_{- \pi/2}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx$$ or $$ I=\frac{1}{2}\int_{- \pi/2}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx=\int_{0}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx.\tag{3}$$ Now use the substitution $u=\frac{\pi}{2}-x$ to get $$ I=\int_{0}^{\pi /2}\frac {\cos^{2008}u}{\sin^{2008}u + \cos^{2008}u} \, du.\tag{4}$$ Now adding (3) and (4) will give you $ 2I=\pi/2$ or $I=\pi/4$.

xpaul
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  • Thanks, I didn't know this trick. How did you learn it, and in general how to learn integration tricks like this one? I can't imagine myself ever have to invent something like this. – user165134 Jul 18 '14 at 20:36
  • @user165134, this trick is very common for some integrals which have some symmetrical properties. – xpaul Jul 18 '14 at 20:38
  • ah! How to learn more tricks? – user165134 Jul 18 '14 at 20:38
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    @user165134, work on more problems. – xpaul Jul 18 '14 at 20:40
  • I mean for example this integration technique you used in your problem I would never have thought of it, then how could I ever learn something if I can never find it? – user165134 Jul 18 '14 at 20:44
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    @user165134 A bit note: $$ \int_{-\Large\frac\pi2}^{\Large\frac\pi2}\frac{\sin^nx}{\sin^nx+\cos^nx}\ dx=2\int_{0}^{\Large\frac\pi2}\frac{\sin^nx}{\sin^nx+\cos^nx}\ dx $$ is only valid for $n$ is even integer. +1 anyway. – Tunk-Fey Jul 19 '14 at 06:19