Feynman's trick IV

Diff under the integral sign

We show the proof for the following result using the Feynman's technique: \[ \int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi \ln(1+\sqrt{2})}{2} \] Proof Consider the intgral \[ \Phi(a)= \int_{0}^{\frac{\pi}{2}} \frac{\arctan a\sin x}{\sin x} dx \] Differentiatinting the parameter $a$ under the integral sign: \begin{align*} \Phi'(a) =\frac{d}{da} \int_{0}^{\frac{\pi}{2}} \frac{\arctan a\sin x}{\sin x} dx =& \int_{0}^{\frac{\pi}{2}} \frac{d}{da} \frac{\arctan a\sin x}{\sin x} dx \\ =& \int_{0}^{\frac{\pi}{2}} \frac{1}{a^2\sin^2{x}+1} dx\\ =& \frac{\arctan \left(\tan x \sqrt{a^2+1} \right)}{\sqrt{a^2+1}}\Big|_{0}^{\frac{\pi}{2}} \\ =& \frac{\pi}{2\sqrt{a^2+1}} \end{align*} Therefore \begin{align*} \Phi(a) =& \int \frac{\pi}{2\sqrt{a^2+1}} da \\ =& \frac{\pi\operatorname{arcsinh} a}{2} + C\\ =& \frac{\pi \ln(a+\sqrt(a^2+1))}{2} + C \end{align*} Now if we evaluate $\displaystyle \Phi(1) $ Then \[\int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi\ln(1+\sqrt(2))}{2} + C \] Finally, if we evaluate $\displaystyle \Phi(0)$ then $\arctan a\sin(x)=0$ and $C=0$ \[\boxed{\int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi\ln(1+\sqrt(2))}{2} } \]