Can anyone evaluate this definite integral? $$I = \int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{\frac{\pi}{2}-x}}+\sin{\sqrt x}}dx$$ 
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!
 A: $$I = \int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{(\frac{\pi}{2}-x)}}+\sin{\sqrt x}}dx$$
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 
$$I+I= \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-x}}}{\sin{\sqrt{(x)}}+\sin{\sqrt {\pi/2-x}}}dx+\int_0^{\pi/2}\frac{\sin{\sqrt x}}{\sin{\sqrt{(\frac{\pi}{2}-x)}}+\sin{\sqrt x}}dx$$
$$2I= \int_0^{\pi/2}\frac{\sin{\sqrt {\pi/2-x}}+\sin(\sqrt x)}{\sin{\sqrt{(x)}}+\sin{\sqrt {\pi/2-x}}}dx=\int_0^{\pi/2}1 \cdot  dx$$
