I have given my friend an integral that I genuinely don't know how to calculate: $$\int_{0}^{\frac{\pi}{2}} \sinh^{-1}(\sin(x))dx$$ He tried differentiating under the integral and complex integration, neither of which worked.

I would love to see a way of calculating this. I know for a fact that it equals Catalan's Constant ~$.915$.

Thanks in advance!

Edit: If anyone is curious where I found this, I found it under the article for Catalan's Constant in Wolfram Alpha.

  • $\begingroup$ Is this a fun question? $\endgroup$ Commented Nov 14, 2017 at 0:11
  • $\begingroup$ Quick: what is $\arcsin(\sin(x))$? No calculus, please. $\endgroup$
    – Lubin
    Commented Nov 14, 2017 at 0:11
  • 1
    $\begingroup$ @TomHimler You gave your friend a problem that you can't solve!? That's a dirty trick! >:D $\endgroup$ Commented Nov 14, 2017 at 0:19
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    $\begingroup$ @TobyMak But... that's an indefinite integral! There are all kinds of tricks for definite integrals that can't be applied to indefinite ones. $\endgroup$ Commented Nov 14, 2017 at 0:22
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    $\begingroup$ There's an explanation of sorts given at the source that is cited in that article. cs.cmu.edu/~adamchik/articles/catalan/catalan.htm $\endgroup$ Commented Nov 14, 2017 at 0:34

1 Answer 1


For any $a\in(0,1)$ we have $$ \int_{0}^{\pi/2}\frac{\sin(x)\,dx}{\sqrt{1+a^2 \sin^2 x}}=\frac{\arctan a}{a} $$ hence by applying $\int_{0}^{1}(\ldots)\,da $ to both sides we get $$\int_{0}^{\pi/2}\operatorname{arcsinh}(\sin x)\,dx=\int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n a^{2n}}{(2n+1)}\,da = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}=G. $$ Differentiation under the integral sign does work.

An alternative way is to notice that the Taylor series of the (hyperbolic) arcsine function and the integrals $\int_{0}^{\pi/2}(\sin x)^{2n+1}\,dx$ interact by simplifying each other: $$\int_{0}^{\pi/2}\operatorname{arcsinh}(\sin x)\,dx=\sum_{n\geq 0}\frac{\binom{2n}{n}(-1)^n}{(2n+1)4^n}\int_{0}^{\pi/2}(\sin x)^{2n+1}\,dx=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}. $$ This is reminiscent of one of Euler's proofs of the Basel problem.


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