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How to evaluate this definite integral

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sin t +\cos t}{\sqrt{\sin 2t}}dt$$

I have tried using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ but not able to solve it.

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1 Answer 1

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Hint

$$\frac{\sin t + \cos t}{\sqrt{\sin 2t}}=\frac{\sin t + \cos t}{\sqrt{2\sin t\cos t}}=-\frac{\left(\cos t-\sin t\right)'}{\sqrt{1-\left(\cos t - \sin t\right)^2}}$$

So you then have:

$$\int \frac{\sin t + \cos t}{\sqrt{\sin 2t}} \,\mbox{d}t=-\int\frac{\left(\cos t-\sin t\right)'}{\sqrt{1-\left(\cos t - \sin t\right)^2}}\,\mbox{d}t = -\arcsin\left(\cos t - \sin t\right) + C$$

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