Evaluate $\int ( \frac{1}{\sin x} - \sin x - \sin x \log(\sin x) ) dx$

Question

This one looks easy but I still could'nt figure it out.

$$\int ( \frac{1}{\sin x} - \sin x - \sin x \log(\sin x) ) dx$$

I tried substituting $\log(\sin x)=z$ but that's not working.Any suggestions?

Answer 1

$$\int -\sin x \log \sin x \, dx=\int (\cos x)' \log \sin x \, dx$$

$$=\cos x \log \sin x - \int \cos x \cdot \frac{\cos x}{\sin x}\, dx$$

Note that $\cos x \cdot \frac{\cos x}{\sin x}=\frac{\cos^2x}{\sin x}=\frac{1-\sin^2x}{\sin x}=\frac{1}{\sin x}-\sin x$, so we have:

$$\int -\sin x \log \sin x \, dx=\cos x \log \sin x + \int (\sin x - \frac{1}{\sin x}) \, dx$$

Bringing this back to the starting integral, we obtain our solution as:

$$\cos x \log \sin x +C$$