Impossible integral? $\int_0^{\pi/2}\frac{\sin^2(x)}{\sqrt{x^2+(\pi/2-x)^2}}$

Question

I have absolutely no idea how to solve:$$\int_0^{\pi/2}\frac{\sin^2(x)}{\sqrt{x^2+(\pi/2-x)^2}}$$I think I am supposed to use trig sub, but I can't seem to put it in the right form?

Answer 1

By the King property$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx=\tfrac12\int_a^b[f(x)+f(a+b-x)]dx,$$your integral is$$\frac12\int_0^{\pi/2}\frac{\sin^2x+\cos^2x}{\sqrt{x^2+(\pi/2-x)^2}}dx=\frac12\int_0^{\pi/2}\frac{dx}{\sqrt{2x^2-\pi x+\pi^2/4}}=\frac{\operatorname{arsinh}1}{\sqrt{2}}.$$