My textbook has the following problem:
$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \ dx$$
Using the trigonometric identity:
$$\sin 2x = 2 \sin x \cos x$$
the integration simplifies to:
$$\frac{1}{\sqrt{2}} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \ dx$$
I'm stuck here.
Best Answer
\begin{align} &\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x +\cos x}{\sqrt{\sin 2x}} \ dx\\ =&\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d(\sin x -\cos x)}{\sqrt{1-(\sin x -\cos x)^2}} \\ =& \ \arcsin(\sin x -\cos x)\bigg|_{\frac{\pi}{6}}^{\frac{\pi}{3}} =2\arcsin\frac{\sqrt3-1}2 \end{align}