Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$. 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?
 A: 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 $I$.
A: 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 $$\int_0^{\frac\pi2}\frac{h(\sin x)}{h(\sin x)+h(\cos x)}dx$$
A: 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 $\pi/4$.
