Evaluate $$\alpha=\int_{\pi \over 6}^{\pi \over 3}\frac{\sin t+\cos t}{\sqrt{\sin 2t}}dt$$
My attempt:
First I tried to substitute ${π \over 2}-x$, but didn't work .
Then I made square in denominator
$$\alpha=\int_{\pi \over 6}^{\pi \over 3}\frac{\sin t+\cos t}{\sqrt{(\sin t+\cos t)^2-1}}dt$$
I could put $\sin t + \cos t =x$ but above is not differentiation of it.
