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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?


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

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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

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 $\pi/4$.


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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