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

$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+\tan^3(x)} dx$$
Then to evaluate $\int \frac{1}{1+\tan^3(x)}dx$, I let $\tan(x)=t$ and hence $dt = \sec^2(x) dx$, which implies $dx=\frac{1}{\sec^2(x)}dt$.
Then I don't know how to continue.
 A: Replacing $x$ by $\pi/2-x$, the integral becomes
$$ I = \int_0^{\pi/2} \frac{\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx. $$
Adding together with the original form of $I$,
$$ 2I = \int_0^{\pi/2} \frac{\sin^3{x}+\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}, $$
so $I = \pi/4$.
This trick works on any integral of the form
$$ \int_0^a \frac{f(x)}{f(x)+f(a-x)} \, dx: $$
the answer is always $a/2$.
A: Let $x=\frac{\pi}2-u$. Then $dx=-du$ and the integral becomes:
$$I=-\int_{\pi/2}^0 \frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du\\=\int_0^{\pi/2}\frac{\cos^3(u)}{\cos^3 (u)+\sin^3(u)}du$$
So your integral satisfies:
$$2I=\int_0^{\pi/2} 1dx=\frac{\pi}2$$
A: Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$ ......(1)
Using property -
$\int_a^{b} f(x) dx = \int_a^{b} f(a + b - x) dx$
We have,
$I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(\frac{\pi}{2}-x)}{\cos^3(\frac{\pi}{2}-x)+\sin^3(\frac{\pi}{2}-x)} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{\cos^3(x)}{\sin^3(x)+\cos^3(x)} dx$ .....(2)
Adding equation (1) and (2),
$2I = \int_0^{\frac{\pi}{2}} dx$
$2I = \left[x\right]_0^{\frac{\pi}{2}}$
$2I = \frac{\pi}{2}$
$I = \frac{\pi}{4}$
A: 
I entered this into Wolfram alpha and this is what it gives me. 
