TITLE: Is there an elegant way to solve $\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$?
QUESTION [8 upvotes]: The integral is:
$$\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$$
I used weierstraß substitution 
$$t:=\tan(\frac{x}{2})$$
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2}{1+t^2}dt$$
Got this:
$$\int \frac{8t^4-8t^2}{t^8−2t^7+2t^6−6t^5−6t^3−2t^2−2t−1}dt$$
and with partial fraction expansion the final answer is:
$$\frac{1}{4}[\ln(\sin(x)+\cos(x))-\cos(x)*(\sin(x)+\cos(x))]+C$$
It is a long way and I am very convinced, there is a shorter way, maybe you know one? Thanks

REPLY [2 votes]: $\sin(x)+\cos(x)=\sqrt2\cos\left(x-\frac\pi4\right)$. Let $u=x-\frac\pi4$, then
$$
\begin{align}
\frac{\sin^2(x)\cos(x)}{\sin(x)+\cos(x)}
&=\frac{(\cos(u)+\sin(u))^2}2\frac{\cos(u)-\sin(u)}{\sqrt2}\frac1{\sqrt2\cos(u)}\\
&=\frac{\cos(u)-\sin(u)+2\sin(u)\cos^2(u)-2\sin^2(u)\cos(u)}{4\cos(u)}\\
&=\frac14\left(1-\frac{\sin(u)}{\cos(u)}+2\sin(u)\cos(u)-2\sin^2(u)\right)\\
&=\frac14\left(\sin(2u)+\cos(2u)-\frac{\sin(u)}{\cos(u)}\right)
\end{align}
$$
Integrating gives
$$
\begin{align}
\int\frac{\sin^2(x)\cos(x)}{\sin(x)+\cos(x)}\mathrm{d}x+C
&=-\frac18\cos(2u)+\frac18\sin(2u)+\frac14\log(\cos(u))+C\\
&=-\frac18\cos\left(2x-\frac\pi2\right)+\frac18\sin\left(2x-\frac\pi2\right)+\frac14\log\left(\cos\left(x-\frac\pi4\right)\right)+C\\
&=-\frac18\sin(2x)-\frac18\cos(2x)+\frac14\log(\cos(x)+\sin(x))+C-\frac{\log(2)}8
\end{align}
$$