Integrate $\int_0^{\frac{\pi}{2}}\frac{\sin^3(x)}{\sin^3(x) + \cos^3(x)}dx$

Question

I am quite sure I saw this somewhere on SE but I couldn't find it. How to evaluate $$\int_0^{\frac{\pi}{2}}\frac{\sin^3(x)}{\sin^3(x) + \cos^3(x)}dx$$

I already have

$$\int \frac{\sin^3(x)}{\sin^3(x) + \cos^3(x)}dx+\int\frac{\cos^3(x)}{\sin^3(x) + \cos^3(x)}dx = \frac{\pi}{2}$$

Ideally I would like to have something like $I_1 + I_2 = \frac{\pi}{2}$ and $I_1 - I_2 = \text{something}$. But

$$\int \frac{\sin^3(x)}{\sin^3(x) + \cos^3(x)}dx-\int\frac{\cos^3(x)}{\sin^3(x) + \cos^3(x)}dx=?$$

I tried to play around with even and odd functions' property. The problem is that $\sin^3(-x) + \cos^3(x) = -\sin^3(x) + \cos^3(x)$.

Apply the substitution $x\mapsto \frac{\pi}{2}-x$. Sum the two equivalent integrals, profit. There is no need to compute any difference. — Jack D'Aurizio, Sep 09 '16 at 04:33
@JackD'Aurizio Oh oh I see. Those trig things are gonna haunt me forever. — 3x89g2, Sep 09 '16 at 04:40

score 0 · Accepted Answer · answered Sep 09 '16 at 04:42

Put $u = \frac \pi 2 - x$. Then $du = -dx$, and $sin(x)=cos(u), $ and $ cos(x)= sin(u).$ For our limits of integration, $x = 0 \to u = \frac \pi 2$, $x= \frac \pi 2 \to u = 0$.

Let $I = \int_0^{\frac \pi 2} \frac {sin^3(x)}{sin^3(x) + cos^3(x)} dx$.

Then $2I= \int_0^{\frac \pi 2} \frac {sin^3(x)}{sin^3(x) + cos^3(x)} dx + \int_{\frac \pi 2}^0 \frac {cos^3(u)}{cos^3(u) + sin^3(u)} (-du)$ $=\int_0^{\frac \pi 2} \frac {sin^3(x)}{sin^3(x) + cos^3(x)} dx + \int_0^\frac \pi 2 \frac {cos^3(u)}{cos^3(u) + sin^3(u)} du$.

But u is a dummy variable. So we have now

$2I = \int_0^{\frac \pi 2} \frac {sin^3(x)}{sin^3(x) + cos^3(x)} dx + \int_0^\frac \pi 2 \frac {cos^3(x)}{cos^3(x) + sin^3(x)} dx = \int_0 ^{\frac \pi 2} \frac {sin^3(x) + cos^3(x)}{sin^3(x) + cos^3(x)}dx = \int_0^{\frac \pi 2} dx = \frac \pi 2.$

Thus $I = \frac 12 * \frac \pi 2 = \frac \pi 4$.

Integrate $\int_0^{\frac{\pi}{2}}\frac{\sin^3(x)}{\sin^3(x) + \cos^3(x)}dx$

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