$$\int_0^π\log(5-4\sin x)dx$$

I tried to proceed with this question using integration by parts but got stuck over $\int_0^π(\frac{-4x\cos x}{5}-4\sin x) dx$.

I also tried to proceed through changing $\cos x$ into the half-angle formula of $\tan \frac x2$ but that $\log$ thing doesn't allow me to make the substitution.

  • $\begingroup$ Taylor series __ $\endgroup$ – EasyTreyballSniper Sep 9 '20 at 4:02
  • $\begingroup$ I didn't get how to use Taylor series in this problem. $\endgroup$ – Kumar Vivek Sep 9 '20 at 4:30
  • $\begingroup$ @KumarVivek he edited and added characters for mathematical formatting. Why are you against it? $\endgroup$ – DatBoi Sep 9 '20 at 4:31
  • $\begingroup$ @KumarVivek I only fixed some MathJax expressions. Actually, you have modified the question, you have just swap the sine and cosine functions. $\endgroup$ – azif00 Sep 9 '20 at 4:39
  • $\begingroup$ We dont change any expression. Even if its wrong, we just format it $\endgroup$ – DatBoi Sep 9 '20 at 5:06

I think that you have some mistakes from the start. Using integartion by parts $$\int \ln(5-4\sin (x))\,dx=x \log (5-4 \sin (x))+4\int\frac{ x \cos (x)}{5-4 \sin (x)}\,dx$$ $$\int_0^\pi \ln(5-4\sin (x))\,dx=\pi \log (5)+4\int_0^\pi\frac{ x \cos (x)}{5-4 \sin (x)}\,dx$$

The last integral is a pure nightmare (try it with this) but it can be evaluated.

By the way $$\int_0^\pi \ln(5-4\sin (x))\,dx=2\int_0^{\frac \pi 2} \ln(5-4\sin (x))\,dx$$ which could approximate (make a plot of it) as the area of the two approximate triangles that is to say $\sim \frac \pi 2 \ln(5) =2.53$ while numerical integration would give $2.41$.

  • $\begingroup$ the answer given in the book is 2πlog2 which doesn't match. $\endgroup$ – Kumar Vivek Sep 9 '20 at 4:46
  • $\begingroup$ @KumarVivek. If we speak about the initial integral, the answer in the book is wrong. $\endgroup$ – Claude Leibovici Sep 9 '20 at 5:04
  • $\begingroup$ will u please explain how to proceed with complex substitution in integrals. $\endgroup$ – Kumar Vivek Sep 9 '20 at 5:20
  • $\begingroup$ @KumarVivek $2\pi\ln\left(2\right) \approx 4.3552$ while your proposed integration is $\approx 2.4063$ which shows that something is wrong with your question. $\endgroup$ – Felix Marin Sep 14 '20 at 1:51

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