> 
> How to compute the following integral?
> $$\int\_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
> 
> 
> 

So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) = \cos(x)$. The integral changes into easier:

$$\int\_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^3(x)+ \cos^3(x)} \, \mathrm{d}x$$

And then I divided by $\cos^3(x)$. It will turn everything to

$$\int\_{0}^{\frac{\pi}{2}} \frac{\sec^2(x)}{\tan^3(x)+1} \, \mathrm{d}x.$$

And I used $u$-substitution setting $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$ and the bounds $u = \tan(\frac{\pi}{2}) = \infty$ and $u =\tan(0) = 0$ and the integral changed into integral to

$$ \int\_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3} $$

This is where I used partial fraction decomposition and my answer is divergent and my answer is wrong:
\begin{align\*}
\int\_{0}^{\infty} \frac{1}{3(u+1)} + \frac{-u+2}{3(u^2-u+1)} \, \mathrm{d}u
\end{align\*}

by this it looks like it will divergent.

The correct answer is $\frac{2\pi}{3\sqrt{3}}$. So, what do I do, next?

Then I did the another method which gives me divergent again,

\begin{align\*}
\int\_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin^3(x)+\cos^3(x)} \, \mathrm{d}x
&= \int\_{0}^{\frac{\pi}{2}} \frac{\tan(x)\sec^2(x)}{1+\tan^3(x)} \, \mathrm{d}x
\end{align\*}

and $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$, the integral

$$\int\_{0}^{\frac{\pi}{2}} \frac{u}{1+u^3} \, \mathrm{d}u$$

\begin{align\*}
\int\_{0}^{\frac{\pi}{2}} \frac{-1}{3(u+1)} + \frac{u+1}{3(u^2-u+1)} \, \mathrm{d}u
\end{align\*}

and did partial fraction and this gives divergent.

I have no idea what to do next for the first method of work, or the second method of work.