0
$\begingroup$

$$\int_0^π\log(5-4\sin x)dx$$

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.

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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 $2.41$.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.