$$I = \int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{\frac{\pi}{2}-x}}+\sin{\sqrt x}}dx$$
All I have is this, "Hint: Let $u = \pi/2-x$." This was meant to be solved in a timely manner and shouldn't be anything a calculus 2 graduate couldn't solve. Thanks!
$$I = \int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{(\frac{\pi}{2}-x)}}+\sin{\sqrt x}}dx$$ let $$u = \pi/2-x$$ thus $$du=-dx $$ also as x is chaging fron 0 to $\pi/2$ u changes from $\pi/2 $ to 0
your integration after substitution becomes ' $$I = \int_{\pi/2}^0\frac{\sin{\sqrt {\pi/2-u}}}{\sin{\sqrt{(u)}}+\sin{\sqrt {\pi/2-u}}}-du$$ $$I = -\int_{\pi/2}^0\frac{\sin{\sqrt {\pi/2-u}}}{\sin{\sqrt{(u)}}+\sin{\sqrt {\pi/2-u}}}du$$ $$I = \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-u}}}{\sin{\sqrt{(u)}}+\sin{\sqrt {\pi/2-u}}}du$$ you can write this integral as $$I = \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-x}}}{\sin{\sqrt{(x)}}+\sin{\sqrt {\pi/2-x}}}dx$$ now add both these integrals above one and this one you should get
$$I+I= \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-x}}}{\sin{\sqrt{(x)}}+\sin{\sqrt {\pi/2-x}}}dx+\int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{(\frac{\pi}{2}-x)}}+\sin{\sqrt x}}dx$$ $$2I= \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-x}}+\sin(\sqrt x)}{\sin{\sqrt{(x)}}+\sin{\sqrt {\pi/2-x}}}dx=\int_0^{\pi/2}1 \cdot dx$$