Evaluate: $I=\int\limits_{0}^{\frac{\pi}{2}}\ln\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}dx$
Evaluate: $$I=\int\limits_{0}^{\frac\pi2}\ln\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}dx$$
First use $\displaystyle \ln a^m=m\ln a$ and $\displaystyle \ln\frac ab=\ln a -\ln b $
Then utilize $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\ \ \ \ (1)$$
so that $$2I=\int_a^b\{f(x)+ f(a+b-x)\}dx$$
to find something really interesting
Again, if $\displaystyle g(x)=\cos x\cdot\ln(1+\sin x)$ what is $g\left(\frac\pi2+0-x\right)?$
So using $(1),I$ should be reduced to $\displaystyle\int_0^1\ln(1+u)du$