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This question already has an answer here:

Evaluate the integral $$\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, \mathrm dx.$$

How can i evaluate this one? Didn't find any clever substitute and integration by parts doesn't lead anywhere (I think).

Any guidelines please?

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marked as duplicate by M.H, Thomas Andrews, apnorton, Ayman Hourieh, Git Gud Jul 9 '13 at 18:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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As $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$$

If $$\begin{eqnarray}I &=& \int^{\frac{\pi}{2}}_0 \frac{\sin^nx}{\sin^nx+\cos^nx} \,dx\\ &=& \int^{\frac{\pi}{2}}_0 \frac{\sin^n\left(\frac\pi2-x\right)}{\sin^n\left(\frac\pi2-x\right)+\cos^n\left(\frac\pi2-x\right)}\, dx\\ &=& \int^{\frac{\pi}{2}}_0 \frac{\cos^nx}{\cos^nx+\sin^nx}\, dx \end{eqnarray}$$

$$\implies I+I=\int_0^{\frac\pi2}dx$$ assuming $\sin^nx+\cos^nx\ne0$ which is true as $0\le x\le \frac\pi2 $

Generalization : $$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx, J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$

$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$

If $a=0,b=\frac\pi2$ and $g(x)=h(\sin x),$

$g(\frac\pi2+0-x)=h(\sin(\frac\pi2+0-x))=h(\cos x)$

So, $J$ becomes $$\int_0^{\frac\pi2}\frac{h(\sin x)}{h(\sin x)+h(\cos x)}dx$$

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    $\begingroup$ I suggested an edit for readability (aligned the equations). I hope you don't mind! $\endgroup$ – Cameron Williams Jul 9 '13 at 17:46
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    $\begingroup$ @CameronWilliams, I added an assumption which you have removed during edit. Let me edit myself $\endgroup$ – lab bhattacharjee Jul 9 '13 at 17:52
  • $\begingroup$ Oops. I think we edited at the same time. :( Sorry about that. $\endgroup$ – Cameron Williams Jul 9 '13 at 17:54
  • $\begingroup$ @CameronWilliams, could you please verify the readability $\endgroup$ – lab bhattacharjee Jul 9 '13 at 18:00
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    $\begingroup$ @Arjang, $$I+I=\int^{\frac \pi2}_0 \frac{\sin^nx+\cos^nx}{\sin^nx+\cos^nx} \,dx=\int_0^\frac\pi21\ dx$$ $\endgroup$ – lab bhattacharjee Dec 27 '14 at 6:46
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Symmetry! This is the same as the integral with $\cos^3 x$ on top.

If that is not obvious from the geometry, make the change of variable $u=\pi/2-x$.

Add them, you get the integral of $1$. So our integral is $\pi/4$.

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Hint: if $$I=\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, dx$$ and $$J=\int^{\frac{\pi}{2}}_0 \frac{\cos^3x}{\sin^3x+\cos^3x}\, dx$$

Then consider $I+J$, and the effect of the substitution $y=\frac{\pi}2-x$ on the integral $I$.

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