Question

integral with log and cos

Answer 1

\[\huge \color{blue} {\int\limits_0^{\pi/2}\frac{ \log(1+\cos \alpha \cos x) }{ \cos x }dx}\]

Answer 2

so far i tried rewriting as
\[\log (1+\cos \alpha \cos x)=\log(1/\alpha+\cos x)+\log \cos \alpha\]
making 2 seperate integrals

Answer 3

i mean\[\frac{\log(1/\cos \alpha+\cos x)}{\cos x}+\frac{\log(\cos \alpha)}{\cos x}\]

Answer 4

i suppose alhpa is a constant,
just for interest sake does the fact that
cos is even help us esp when we substitute\[x=-y\]

Answer 5

so maybe the problem to be post is
\[\int\limits \frac{\ln(a+\cos x)}{\cos x}dx\]
since the send part is elementary

Answer 6

i don't think this is that straight forward
maple gives me an ungodly mess, and wolfram won't do it at all

Answer 7

i will re check it if theres any error in my input

Answer 8

Maybe the first step is:
\[\int\limits_{0}^{\frac{ \pi }{ 2 }}\frac{ \log(1+\cos \alpha cosx) }{ cosx }dx = \lim_{t \rightarrow \frac{ \pi }{ 2 }}\int\limits_{0}^{t}\frac{ \log(1+\cos \alpha cosx) }{ cosx }dx\]
Since cosx = 0 when x = pi/2. I'm still trying to get rid of that log at the top...

Answer 9

so i realize that these question is advanced for me its called alpha is called a parameter and somehow it eds up beign the variable
howerver i have the right hand side
explanation invloves integral with dummy variable x and then later alpha in will post solution soon