Why $\int{\log^2(2\sin(\pi x))}dx\neq\frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$? Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:
$$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$

UPDATE:
Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:
$$\int{\log^2(2\sin(\pi x))}dx = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$

Let $f(x)=2\sin(\pi x)$. 
Then, 
$$\frac{d \log^2(f(x))}{dx}=2\log f(x)\times (\log f(x))^\prime.$$
Here, $$(\log f(x))^\prime=\frac{f^\prime (x)}{f(x)}=\frac{2\cos(\pi x)\times \pi}{2\sin(\pi x)}.$$
As a result, we have
$$\frac{d \log^2(f(x))}{dx}=2\log (2\sin(\pi x))\times \frac{2\cos(\pi x)\times \pi}{2\sin(\pi x)}.$$
If you feel any difficulty in my answer, just let me know.
