# definite integration of logarithmic function

$$\int_0^π\log(5-4\sin x)dx$$

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.

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

I think that you have some mistakes from the start. Using integartion by parts $$\int \ln(5-4\sin (x))\,dx=x \log (5-4 \sin (x))+4\int\frac{ x \cos (x)}{5-4 \sin (x)}\,dx$$ $$\int_0^\pi \ln(5-4\sin (x))\,dx=\pi \log (5)+4\int_0^\pi\frac{ x \cos (x)}{5-4 \sin (x)}\,dx$$
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 $$2.41$$.
• @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. – Felix Marin Sep 14 '20 at 1:51