Question

honestly, I did make up this problem, but I just want to see if I am able to work it out.

Answer 1

\[\int\limits_{ }^{ }\ln(\cos v)~dv\]

Answer 2

or maybe I can do this:
\[\int\limits\limits_{ }^{ }\ln(\cos v)~dv = \int\limits\limits_{ }^{ }\frac{\sin v~\ln(\cos v)}{\sin v}~dv \]\[\color{blue}{u=\cos v,~~~~~-du=\sin v~dv,}\]\[u=\cos v~~~~\rightarrow~~~~u^2=\cos^2v~~~~\rightarrow~~~~u^2-1=\cos^2v-1\]\[~~~~\rightarrow~~~~~1-u^2=\sin^2v~~~~~\rightarrow~~~~~\color{blue}{\sin v=\sqrt{1-u^2}}\]

Answer 3

\[\int\limits_{ }^{ }\frac{\ln(u)}{\sqrt{1-u^2}}~du\]

Answer 4

\[\int\limits_{ }^{ } \frac{\ln u}{\sqrt{1-u^2}}~du\]\[\int\limits_{ }^{ } \frac{u~\ln u}{u \sqrt{1-u^2} }~du\]\[\color{blue}{s=1-u^2,~~~~~~~~-\frac{1}{2}du=u~du}\]\[u=\sqrt{1-s}\]
and we have so far, that\[\cos(v)=u=\sqrt{1-s}\]\[\cos^2(v)=u^2=1-s\]\[\cos^2(v)-1=u^2-1=-s\]\[\color{red}{\sin^2(v)-1}=1-u^2=\color{red}{s}\]

Answer 5

forgot,
\[\ln(u)=\frac{1}{2}\ln (1-s)\]

Answer 6

\[-\frac{1}{4}\int\limits_{ }^{ } \frac{\ln(1-s)}{s \sqrt{1-s}}ds\]

Answer 7

the second blue, should be, btw, -1/2 ds = u du

Answer 8

\[p=1-s\]\[s=1-p\]\[-dp=ds\]

Answer 9

the last red, I wrote should be sin^2v=s without -1 next to sin^2v

Answer 10

for the current sub I have:
\[\sin^2(v)=s=1-p\]\[\sin^2(v)-1=s-1=-p\]\[\cos^2(v)=p\]

Answer 11

\[\frac{1}{4} \int\limits_{ }^{ } \frac{\ln p }{(1-p)\sqrt{p}}~dp\]

Answer 12

Very interesting. Does wolfram produce an answer

Answer 13

http://www.wolframalpha.com/input/?i=integral+of+ln%28cos+x%29
I am like, "my make up is not very good-:( "

Answer 14

maybe there will be a way to writewe it using a sigma notation, but will wait for the wise man:)