How do I evaluate this integral $I = \int_{0}^{2 \pi} \ln (\sin x +\sqrt{1+\sin^2 x}) dx$?

Question

I used some variables change to evaluate this integral but i'm not succeed may I have some wrong step as trigono-transformation.Then Is there some one who can show me how do evaluate this : $$I = \int_{0}^{2 \pi} \ln (\sin x +\sqrt{1+\sin^2 x}) dx$$

Thank you for any help .

Answer 1

In the First : $\sinh x = \frac{e^x - e^{-x} }{2} \Longrightarrow \sinh^{-1}x = \ln(x+\sqrt{x^2+1}).$

to get this just solve the equation $y=\sinh x$ to get the above inverse function (notice that $e^x>0$). The integral becomes :

\begin{align*} \int_0^{2\pi} \sinh^{-1} \sin x\ \mathrm{d}x &= \int_0^{\pi} \sinh ^{-1} \sin x \ \mathrm{d}x + \int_{\pi}^{2\pi} \sinh^{-1} \sin x \ \mathrm{d}x \\ &= \int_0^{\pi} \sinh ^{-1} \sin x \ \mathrm{d}x+ \int_0^{\pi} \sinh ^{-1} \sin (x+\pi) \ \mathrm{d}x \\ &= \int_0^{\pi} \sinh ^{-1} \sin x \ \mathrm{d}x - \int_0^{\pi} \sinh ^{-1} \sin x \ \mathrm{d}x \\ &=0.\end{align*}

(the $\sinh^{-1}$ function is odd)

Answer 2

Here is another way of evaluating the integral. Let $f(x)$ be an odd function with period $2\pi$. Then, we will show that

$$\int_0^{2\pi}\log \left(f(x)+\sqrt{f^2(x)+1}\right)dx=0$$

$$\begin{align} \int_0^{2\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx&=\int_{-\pi}^{\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx\\\\ &=\int_{-\pi}^{0}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx\\ &+\int_{0}^{\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx\\\\ &=\int_{0}^{\pi}\log\left(-f(x) +\sqrt{1+f^2(x)}\right)dx\\ &+\int_{0}^{\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx\\\\ &=\int_0^{\pi}\left(\log(-f(x) +\sqrt{1+f^2(x)}\right)+\log\left(f(x) +\sqrt{1+f^2(x)})\right)dx\\\\ &=\int_0^{\pi}\log (1) dx\\\\ &=0 \end{align}$$

Inasmuch as $\sin x$ is odd with period $2\pi$, we immediately have

$$\bbox[5px,border:2px solid #C0A000]{\int_0^{2\pi}\log \left(\sin x+\sqrt{\sin^2 x+1}\right)dx=0}$$