excluderho

Answered

2022-06-26

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

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

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

UPDATE:

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

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

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

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

UPDATE:

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

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

Answer & Explanation

Hadley Cunningham

Expert

2022-06-27Added 20 answers

Using Chain rule:

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

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

polivijuye

Expert

2022-06-28Added 16 answers

Let $f(x)=2\mathrm{sin}(\pi x)$

Then,

$\frac{d{\mathrm{log}}^{2}(f(x))}{dx}=2\mathrm{log}f(x)\times (\mathrm{log}f(x){)}^{\mathrm{\prime}}.$

Here,

$(\mathrm{log}f(x){)}^{\mathrm{\prime}}=\frac{{f}^{\mathrm{\prime}}(x)}{f(x)}=\frac{2\mathrm{cos}(\pi x)\times \pi}{2\mathrm{sin}(\pi x)}.$

As a result, we have

$\frac{d{\mathrm{log}}^{2}(f(x))}{dx}=2\mathrm{log}(2\mathrm{sin}(\pi x))\times \frac{2\mathrm{cos}(\pi x)\times \pi}{2\mathrm{sin}(\pi x)}.$

If you feel any difficulty in my answer, just let me know.

Then,

$\frac{d{\mathrm{log}}^{2}(f(x))}{dx}=2\mathrm{log}f(x)\times (\mathrm{log}f(x){)}^{\mathrm{\prime}}.$

Here,

$(\mathrm{log}f(x){)}^{\mathrm{\prime}}=\frac{{f}^{\mathrm{\prime}}(x)}{f(x)}=\frac{2\mathrm{cos}(\pi x)\times \pi}{2\mathrm{sin}(\pi x)}.$

As a result, we have

$\frac{d{\mathrm{log}}^{2}(f(x))}{dx}=2\mathrm{log}(2\mathrm{sin}(\pi x))\times \frac{2\mathrm{cos}(\pi x)\times \pi}{2\mathrm{sin}(\pi x)}.$

If you feel any difficulty in my answer, just let me know.

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