$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
 A: Put $$I =  \displaystyle\int{\dfrac{\sin^3 x}{\cos^3 x  +\sin^3 x }}\mathrm{d}x , \quad J = \displaystyle\int{\dfrac{\cos^3 x}{\cos^3 x  +\sin^3 x }}\mathrm{d}x.$$
We have $$I + J = \displaystyle\int{\mathrm{d}x} = x+ C.$$
and
\begin{equation*}
I - J =  \displaystyle\int{\dfrac{\sin^3 x - \cos^3 x}{\cos^3 x  +\sin^3 x }}\mathrm{d}x =  \displaystyle\int{\dfrac{(\sin x - \cos x)(1 + \sin x \cdot \cos x)}{(\sin x + \cos x)(1 - \sin x \cdot \cos x)}\mathrm{d}x}
\end{equation*}
Put $t = \sin x + \cos x$,  then $\mathrm{d}t = -(\sin x - \cos x) \mathrm{d}x$ and $ \sin x \cdot \cos x = \dfrac{ t^2-1}{2}.$
We get
\begin{equation*}
I - J = \displaystyle\int{\dfrac{t^2 + 1}{t(t^2-3)}\mathrm{d}t} = \dfrac{2}{3}\ln{(t^2-3)}-\dfrac{1}{3}\ln t + C'.
\end{equation*}
and then
\begin{equation*}
I - J =  \dfrac{2}{3}\ln{((\sin x + \cos x)^2-3)}-\dfrac{1}{3}\ln (\sin x + \cos x) + C'.
\end{equation*}
From $I + J $ and $I - J$, we can calculate $I$.
A: If all else fails, the Weierstrass substitution will do it.
A: 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.
A: Well, I'm still not seeing any nice ways of doing it.  I do see at least one way of proceeding though.  First, divide top and bottom by $\cos^3x$
$$\int\frac{\tan^3xdx}{1+\tan^3x}$$
Now make the substitution
$$x=\tan^{-1}u,dx=\frac{du}{1+u^2}$$
$$\int\frac{u^3du}{(1+u^2)(1+u)(1-u+u^2)}$$
at which point it can be solved by partial fractions.
