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.
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.