Question

integral log [1-sin x/ 1+sin x]dx

Answer 1

Let S=∫π0log(1+sin(x))dx

cos(x)=sin(π2−x)

then S=∫π2−π2log(1+cos(x))dx

log(1+cos(x))=log(eix+2+e−ix2)=2log(eix2+e−ix2)−log(2)

=2[ix2+log(1+e−ix)]−log(2)=log(1+eix)+log(1+e−ix)−log(2)

=∑∞n=1(−1)n−1einxn+∑∞n=1(−1)n−1e−inxn−log(2)

=2∑∞n=1(−1)n−1cos(nx)n−log(2)

Now integrate

S=∫π2−pi2log(1+cos(x))dx=2∑∞n=1(−1)n−1n∫π2−π2cos(nx)dx−πlog(2)

First

∑∞n=1(−1)n−1n∫π20cos(kx)dx=∑∞n=1(−1)n−1n2sin(kx2)

=∑∞n=1(−1)n(2n+1)n=G

Here G is Catalans constant

Then by symmetry

2∑∞n=1(−1)n−1n∫π2−π2=4G

∴S=4G−πlog(2)

Then S=4G−πlog(2)≈1.4862