[Math] $\int_{\pi/6}^{\pi/3} (\sin x + \cos x) / \sqrt{\sin(2x)}\, dx$

Question

Question How to solve the following integral?
$$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{\sqrt{\sin(2x)}}\,dx$$

Attempt Using the property of definite integrals and putting $\pi/6 + \pi/3 – x$ in place of $x$ did not help. So I tried simplifying the function and I got $[(\tan x)^{1/2} + (\cot x)^{1/2}]/2^{1/2}$, with which I don't know how to proceed?

Any helps appreciated.

Answer 1

$$\sin{2x}=1-(\sin{x}-cos{x})^2$$

$$\int{\frac{(\sin{x}+\cos{x})dx}{\sqrt{1-(\sin{x}-\cos{x})^2}}}$$

Let,

$$\sin{x}-\cos{x}=t$$ $$\frac{dt}{dx}=\sin{x}+\cos{x}$$

$$\int \frac{dt}{\sqrt{1-t^2}}$$

$$=\arcsin{t}=\arcsin{(\sin{x}-\cos{x})}$$

Putting limits,

$$=\arcsin{\frac{\sqrt{3}-1}{2}}-\arcsin{\frac{1-\sqrt{3}}{2}}$$

As a sidenote, I have always been taught to substitute $$\sin{2x}=1+(\sin{x}+\cos{x})^2$$ when there are $(\sin{x}-\cos{x})$ and $\sin{2x}$ in the same integral. $$\sin{2x}=(\sin{x}-\cos{x})^2-1$$ when there are $(\sin{x}+\cos{x})$ and $\sin{2x}$ in the same integral. Although exceptions may exist.