Question

Integrate log(1+cosx)dx from 0 to pi

Answer 1

\[\Huge \int\limits_{0}^{\pi} \log(1+cosx)\]

Answer 2

This requires a bit of trick. Hint:
\[\int\limits_{a}^{b}\log_{n}x = (x \log_{n}x-x/\ln(n))+C \]

Answer 3

I am at
\[\Huge I=\int\limits_{0}^{\pi}logsinx\]

Answer 4

dunno what to do next

Answer 5

Well do you know how to integrate by parts? If you don't just post here and I will help you solve this step by step.

Answer 6

\[ 1 + \cos (x) = 2 \cos^2(x/2)\]
make use of log function and change of variables. If you know how to integrate log(sin(x)) from 0 to pi then you should know this too.

Answer 7

\[ I = \int_0^\pi \log(\sin(x))dx = \log(2) \pi + \int_0^\pi \log(\cos(x/2))dx + \int_0^\pi \log(\sin(x/2))dx \\
= \pi \log(2) + 2 \int_0^{\pi/2} \log(\cos(x))dx + 2 \int_0^{\pi/2} \log(\sin(x))dx \\
= \pi \log(2) + 2 \left (\int_0^{\pi/2} \log(\cos(x))dx + \int_0^{\pi/2} \log(\sin(x))dx \right ) \\=
\pi \log(2) + 2 \left (\int_0^{\pi/2} \log(\sin(x+\pi/2))dx + \int_0^{\pi/2} \log(\sin(x))dx \right )
\\=\pi \log(2) + 2 \left (\int_{\pi/2}^\pi \log(\sin(x))dx + \int_0^{\pi/2} \log(\sin(x))dx \right )
\\=\pi \log(2) + 2I
\]

Answer 8

there are more methods
http://math.stackexchange.com/questions/354795/evaluate-int-0-pi-ln-left-sin-theta-right-d-theta