# Evaluate $\int _{\frac{\pi}{6} }^ \frac{\pi}{3} \frac{\sin t+\cos t }{\sqrt{\sin 2t}} dt$ [duplicate]

I want to evaluate $$\int _{\frac{\pi}{6} }^ \frac{\pi}{3} \frac{\sin t+\cos t }{\sqrt{\sin 2t}} dt$$ I tried to use the properties of the integral to reduce the integral. But I failed. Also I tried using $$\sin C+ \sin D$$ and $$\sin 2t$$ trignometric identities. Nothing worked!!

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But that answer is not satisfactory for me because When a person see that integral for the first time , how do he know about that particular trick. Is there any elegant way to solve this problem?

## marked as duplicate by zahbaz, mrtaurho, Lee David Chung Lin, José Carlos Santos, ancientmathematicianFeb 6 at 12:33

• After using $\sin 2t = 2 \sin t \cos t$, your integral is equal to $\frac{1}{\sqrt{2}}\int_{\pi/6}^{\pi/3} \left(\sqrt{\tan t} + \sqrt{\cot t}\right)dt$. Check out this post, specifically robjohn's answer. – zahbaz Feb 6 at 6:42
Substitute $$u=\cos t-\sin t$$, so that $$du=-(\sin t+\cos t)dt$$: $$\int_{\pi/6}^{\pi/3}\frac{\sin t+\cos t}{\sqrt{\sin 2t}}dt =\int_{-\alpha}^\alpha\frac{1}{\sqrt{1-u^2}}du=[\arcsin(u)]_{-\alpha}^\alpha=2\arcsin(\frac{\sqrt3-1}{2})$$
Note that $$1-u^2=\sin(2t)$$ and $$\alpha=(\sqrt3-1)/2$$.