charlygyloavao9

2022-09-29

Definite integration of logarithmic function

${\int}_{0}^{\pi}\mathrm{log}(5-4\mathrm{sin}x)dx$

I tried to proceed with this question using integration by parts but got stuck over

${\int}_{0}^{\pi}(\frac{-4x\mathrm{cos}x}{5}-4\mathrm{sin}x)dx$

${\int}_{0}^{\pi}\mathrm{log}(5-4\mathrm{sin}x)dx$

I tried to proceed with this question using integration by parts but got stuck over

${\int}_{0}^{\pi}(\frac{-4x\mathrm{cos}x}{5}-4\mathrm{sin}x)dx$

Lohre1x

Beginner2022-09-30Added 8 answers

I think that you have some mistakes from the start. Using integartion by parts

$\int \mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=x\mathrm{log}(5-4\mathrm{sin}(x))+4\int \frac{x\mathrm{cos}(x)}{5-4\mathrm{sin}(x)}\phantom{\rule{thinmathspace}{0ex}}dx$

${\int}_{0}^{\pi}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=\pi \mathrm{log}(5)+4{\int}_{0}^{\pi}\frac{x\mathrm{cos}(x)}{5-4\mathrm{sin}(x)}\phantom{\rule{thinmathspace}{0ex}}dx$

By the way

${\int}_{0}^{\pi}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=2{\int}_{0}^{\frac{\pi}{2}}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx$

which could approximate (make a plot of it) as the area of the two approximate triangles that is to say $\sim \frac{\pi}{2}\mathrm{ln}(5)=2.53$ while numerical integration would give $2.41$

$\int \mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=x\mathrm{log}(5-4\mathrm{sin}(x))+4\int \frac{x\mathrm{cos}(x)}{5-4\mathrm{sin}(x)}\phantom{\rule{thinmathspace}{0ex}}dx$

${\int}_{0}^{\pi}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=\pi \mathrm{log}(5)+4{\int}_{0}^{\pi}\frac{x\mathrm{cos}(x)}{5-4\mathrm{sin}(x)}\phantom{\rule{thinmathspace}{0ex}}dx$

By the way

${\int}_{0}^{\pi}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx=2{\int}_{0}^{\frac{\pi}{2}}\mathrm{ln}(5-4\mathrm{sin}(x))\phantom{\rule{thinmathspace}{0ex}}dx$

which could approximate (make a plot of it) as the area of the two approximate triangles that is to say $\sim \frac{\pi}{2}\mathrm{ln}(5)=2.53$ while numerical integration would give $2.41$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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