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I want to evaluate the integral: $$I=\int_{0}^{\pi/2}\ln \left ( \frac{(1+\sin x)^{1+\cos x}}{1+\cos x} \right )\,dx$$
Well, the sub $u=\pi/2-x$ does not give me any result. In fact it makes the integral more complicated that it actually is, unless I do not see something.
The method above is the only one I used since I do not see something else in this point. Any help would be grateful.
The idea in the following is to simplify using logarithmic identities, and then to get rid of the nasty integration term using the symmetry of the limits. $$\begin{align} I&=\int_{0}^{\Large\frac\pi2}\ln \left ( \frac{(1+\sin x)^{1+\cos x}}{1+\cos x} \right )\,dx \\&=\int_{0}^{\Large\frac\pi2}[(1+\cos x)\ln(1+\sin x)-\ln({1+\cos x})]\,dx \\&=\int_{0}^{\Large\frac\pi2}[\cos x\ln (1+\sin x)+\ln(1+\sin x)-\ln({1+\cos x})]\,dx \\&=\int_{0}^{\Large\frac\pi2}\cos x\ln (1+\sin x)\,dx+\int_{0}^{\Large\frac\pi2}[\ln(1+\sin x)-\ln({1+\cos x})]\,dx \\&=\int_{0}^{\Large\frac\pi2}\cos x\ln (1+\sin x)\,dx \end{align}$$ in which the second term vanished because $$\int_{0}^{\Large\frac\pi2}\ln(1+\sin x)\,dx=\int_{0}^{\Large\frac\pi2}\ln(1+\cos x)\,dx$$
Now performing the substitution $u=1+\sin x$, we get
$$\begin{align} I&=\int_{0}^{\Large\frac\pi2}\cos x\ln (1+\sin x)\,dx \\&=\int_{1}^{2}\ln u\,du \\&=\bigg[u\ln u-u\bigg]_1^2 \\&=2\ln2-1. \end{align}$$
