excluderho

2022-06-26

Why $\int {\mathrm{log}}^{2}\left(2\mathrm{sin}\left(\pi x\right)\right)dx\ne \frac{{\mathrm{log}}^{3}\left(2\mathrm{sin}\left(\pi x\right)\right)\mathrm{sin}\left(\pi x\right)}{3\pi \mathrm{cos}\left(\pi x\right)}$?
Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:
$\frac{d{\mathrm{log}}^{2}\left(2\mathrm{sin}\left(\pi x\right)\right)}{dx}=\frac{{\mathrm{log}}^{3}\left(2\mathrm{sin}\left(\pi x\right)\right)\mathrm{sin}\left(\pi x\right)}{3\pi \mathrm{cos}\left(\pi x\right)}$
UPDATE:
Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:
$\int {\mathrm{log}}^{2}\left(2\mathrm{sin}\left(\pi x\right)\right)dx=\frac{{\mathrm{log}}^{3}\left(2\mathrm{sin}\left(\pi x\right)\right)\mathrm{sin}\left(\pi x\right)}{3\pi \mathrm{cos}\left(\pi x\right)}$

Expert

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

polivijuye

Expert

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

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