Why $\int{\log^2(2\sin(\pi x))}dx\neq\frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$?

Question

Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:

$$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$

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UPDATE:

Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:

$$\int{\log^2(2\sin(\pi x))}dx = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$

Answer 1

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.

Answer 2

Using Chain rule:$$\frac{d }{dx}\log^2(2\sin(\pi x))=2\log(2\sin(\pi x))\times[2\times\pi\times\cos(\pi x)]\times\frac{1}{2\sin(\pi x)}$$

Answer 3

$\frac{d \log^2(2\sin(\pi x))}{dx}=2\log (2\sin(\pi x))\frac{d \log(2\sin(\pi x))}{dx}=-2 \pi \log (2\sin(\pi x)) \cot( \pi x)$