charlygyloavao9

2022-09-29

Definite integration of logarithmic function
${\int }_{0}^{\pi }\mathrm{log}\left(5-4\mathrm{sin}x\right)dx$
I tried to proceed with this question using integration by parts but got stuck over
${\int }_{0}^{\pi }\left(\frac{-4x\mathrm{cos}x}{5}-4\mathrm{sin}x\right)dx$

Lohre1x

I think that you have some mistakes from the start. Using integartion by parts
$\int \mathrm{ln}\left(5-4\mathrm{sin}\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}dx=x\mathrm{log}\left(5-4\mathrm{sin}\left(x\right)\right)+4\int \frac{x\mathrm{cos}\left(x\right)}{5-4\mathrm{sin}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}dx$
${\int }_{0}^{\pi }\mathrm{ln}\left(5-4\mathrm{sin}\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}dx=\pi \mathrm{log}\left(5\right)+4{\int }_{0}^{\pi }\frac{x\mathrm{cos}\left(x\right)}{5-4\mathrm{sin}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}dx$
By the way
${\int }_{0}^{\pi }\mathrm{ln}\left(5-4\mathrm{sin}\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}dx=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(5-4\mathrm{sin}\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}dx$
which could approximate (make a plot of it) as the area of the two approximate triangles that is to say $\sim \frac{\pi }{2}\mathrm{ln}\left(5\right)=2.53$ while numerical integration would give $2.41$

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