<s>[INST] $\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
 [/INST] I think that if this exercise can be solved in a simple fashion (without heavy
computation), then this should be the way to approach it (otherwise, it is
just a mindless computation which just requires to apply some algorithm like
the, so-called, Weierstrass substitution and teaches you nothing).
So, since
$$
\sin^{3}x+\cos^{3}x=\left(  \sin x+\cos x\right)  \left(  \sin^{2}x-\sin x\cos
x+\cos^{2}x\right)  =\left(  \sin x+\cos x\right)  \left(  1-\sin x\cos
x\right)  ,
$$
then we try to express $\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}$ as follows (if
possible), in order to be able to (easily) compute ${\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x$:
\begin{align*}
\frac{\sin^{3}x}{\sin^{3}x+\cos^{3}x} &  =A+B\cdot\frac{\left(  \sin x+\cos
x\right)  ^{\prime}}{\sin x+\cos x}+C\cdot\frac{\left(  1-\sin x\cos x\right)
^{\prime}}{1-\sin x\cos x}=\\
&  =A+B\cdot\frac{\cos x-\sin x}{\sin x+\cos x}+C\cdot\frac{\sin^{2}x-\cos
^{2}x}{1-\sin x\cos x}
\end{align*}
From here we obtain that
$$
\sin^{3}x=A\left(  \sin^{3}x+\cos^{3}x\right)  +B\left(  \cos x-\sin x\right)
\left(  1-\sin x\cos x\right)  +C\left(  \sin x+\cos x\right)  \left(
\sin^{2}x-\cos^{2}x\right)
$$
hence
\begin{align*}
0  & =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left(  \cos x-\sin x\right)
-(B+C)\sin x\cos^{2}x+(B+C)\sin^{2}x\cos x\\
& =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left(  \cos x-\sin x\right)  -(B+C)\sin
x(1-\sin^{2}x)+(B+C)(1-\cos^{2}x)\cos x\\
& =(A+B+2C-1)\sin^{3}x+(A-B-2C)\cos^{3}x+(2B+C)\left(  \cos x-\sin x\right)
\end{align*}
so
$$
\left\{
\begin{array}
[c]{r}
A+B+2C=1\\
A-B-2C=0\\
2B+C=0
\end{array}
\right.
$$
which has the (unique) solution
$$
\left\{
\begin{array}
[c]{l}
A=\frac{1}{2}\\
B=-\frac{1}{6}\\
C=\frac{1}{3}
\end{array}
\right.
$$
hence
\begin{align*}
{\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x &=Ax+B\log\left\vert
\sin x+\cos x\right\vert +C\log\left\vert 1-\sin x\cos x\right\vert
+\text{some constant}\\
&=\frac{x}{2}-\frac{\log\left\vert \sin x+\cos x\right\vert }{6}+\frac
{\log\left\vert 1-\sin x\cos x\right\vert }{3}+\text{some constant}
\end{align*}
Let's hope I didn't make any mistake in my calculations.
</s>