Home » Uncategorized » A logarithmic sin integral

A logarithmic sin integral

Evaluate the integral

    \[\mathcal{J} = \int_{0}^{2\pi} \ln \left ( \sin x + \sqrt{1+\sin^2 x} \right ) \, \mathrm{d}x\]

Solution

1st solution: Recall that \sinh^{-1} x =\ln \left ( x + \sqrt{1+x^2} \right ) \; , \; x \in \mathbb{R}. Thus,

    \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*}

2nd solution: In general , let f be an odd and periodic function of period \mathrm{T}=2\pi. Then,

    \begin{align*} \int_0^{2\pi} \ln \left(f(x) +\sqrt{1+f^2(x)}\right)\;\mathrm{d}x&=\int_{-\pi}^{\pi}\ln \left(f(x) +\sqrt{1+f^2(x)}\right) \;\mathrm{d}x\\ &=\int_{-\pi}^{0}\ln \left(f(x) +\sqrt{1+f^2(x)}\right) \; \mathrm{d}x +\\ &\quad \quad \quad +\int_{0}^{\pi}\ln \left(f(x) +\sqrt{1+f^2(x)}\right)\; \mathrm{d}x\\ &=\int_{0}^{\pi}\ln \left(-f(x) +\sqrt{1+f^2(x)}\right) \; \mathrm{d}x +\\ &\quad \quad \quad + \int_{0}^{\pi}\ln \left(f(x) +\sqrt{1+f^2(x)}\right) \; \mathrm{d}x\\ &=\int_0^{\pi}\bigg(\ln (-f(x) +\sqrt{1+f^2(x)}\right)+\\ &\quad \quad \quad +\ln \left(f(x) +\sqrt{1+f^2(x)} \right) \bigg) \; \mathrm{d}x\\ &=\int_0^{\pi}\ln 1 \; \mathrm{d}x\\ &=0 \end{align*}

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network