How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$? 
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.
 A: Note
\begin{align}
 \int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3} 
&=\frac13
\int_{0}^{\infty} \left(\frac{1}{1+u}+ \frac{2-u}{u^2-u+1}\right)du\\
&= \frac13\int_{0}^{\infty} \left(\frac{1}{1+u}-\frac12 \frac{2u-1}{u^2-u+1}+ \frac32\frac1{(u-\frac12)^2+\frac34} \right)du\\
&=\frac13 \left( \ln\frac{u+1}{\sqrt{u^2-u+1}}+\frac1{\sqrt3}\tan^{-1}\frac{2u-1}{\sqrt3}\right)\bigg|_0^\infty=\frac{2\pi}{3\sqrt3}
\end{align}
A: As proved here
For all $n\geq 2$ we shall show that

$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname{csc}\left(\frac{(2\pi(n-k-1)}{2n-1}\right)$$. Just set $n=2$ we have $$I(1)=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin^3x+\cos^3x}dx =\frac{2\pi}{3\sqrt 3}$$.


Call original integral as $I_1$ After we were done with $x=\frac{\pi}{2}-x$ we have $$I_1=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin^3x +\cos^3x }dx $$ adding the $I_1$ and $I_2$ we have $$I= \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\sin x+\cos x }{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\sec^2x}{\tan^2x-\tan x +1}dx\underbrace {=}_{\tan x=y} \frac{1}{2}\int_0^{\infty}\frac{dy}{y^2-y+1}=\frac{1}{\sqrt 3} \tan^{-1}\left(\frac{2y-1}{\sqrt3}\right)\bigg|_0^{\infty}=\frac{2\pi}{3\sqrt 3} $$
A: Your first method, setting $u=\tan x$, is the substitution suggested by Bioche's rules  Your problem is that there's an error in you computation. You should obtain the integral
$$\int_0^{\infty}\frac{\color{red}u\,\mathrm du}{1+u^3},$$
which is convergent since $\:\frac u{1+u^3}\sim_\infty\frac1{u^2}$.
