# Can we express the following in a closed form? [duplicate]

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.

• Not a very nice one. The antiderivative brought to you by wolfram alpha Jul 28, 2014 at 22:50
• Mathematica returns the simple result of $\ln 4-1$ for this definite integral (though Wolfram Alpha sadly only gives the numerical result). So there's an endpoint. That said, how to get there isn't immediately obvious to me. Jul 28, 2014 at 22:51
• @recursiverecursion Apparently the antiderivative is not elementary... , but something tells me that the definite integral can be evaluated elementary.. Jul 28, 2014 at 22:53
• @Semiclassical What is the result that mathematica returns? Jul 28, 2014 at 22:54

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}