# 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.

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## marked as duplicate by lab bhattacharjee, dragon, Yes, RecklessReckoner, mezJul 29 '14 at 6:17

Not a very nice one. The antiderivative brought to you by wolfram alpha –  recursive recursion Jul 28 '14 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. –  Semiclassical Jul 28 '14 at 22:51
@recursiverecursion Apparently the antiderivative is not elementary... , but something tells me that the definite integral can be evaluated elementary.. –  Tolaso Jul 28 '14 at 22:53
@Semiclassical What is the result that mathematica returns? –  Tolaso Jul 28 '14 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}