Calculating $\int\sin(x)\ln(\sin(x))\, \mathrm{d}x$ I was solving indefinite integrals for preparing me to a math test but I found a very hard one I'm not able to solve. It is
$$\int \sin(x)\ln(\sin(x))\, \mathrm{d}x$$
I tried to solve it by substitution but I wasn't able to express the $dx$ with a smart method. The only way I can think to solve it is by parts but it would be too much difficult in my opinion.
I hope you can help me even with an advice, thanks in advance.

In this particular case you can set $u=\cos(x)$ to get rid of $\sin(x)\mathop{dx}=-\mathop{du}$
Then use $\sin(x)=\sqrt{1-u^2}$ and the logarithm will get rid of the square root and split nicely the product $(1-u)(1+u)$.
It becomes $$\int -\frac 12\ln(1-u^2)\mathop{du}=-\frac 12\left(\int \ln(1+u)\mathop{du}+\int\ln(1-u)\mathop{du}\right)$$
Then use antiderivative of $\int\ln(u)=u\ln(u)-u$
