<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] 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 $I$.
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