Question

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{sinx+cosx}{\sqrt{sin2x}}dx$

Answer 1

We have,

${\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{\sin x+\cos x}{\sqrt{1-1+\sin 2x}}dx$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{\sin x+\cos x}{\sqrt{1-{(\sin x-\cos x)}^2}}dx$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{\frac{d}{dx}(\sin x-\cos x)}{\sqrt{1-{(\sin x-\cos x)}^2}}dx\therefore \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}x$

$={{\left[\sqrt{2}{\sin }^{-1}(\sin x-\cos x)\right]}_{-\frac{\pi }{6}}}^{\frac{\pi }{6}}$

$=\left[\sqrt{2}{\sin }^{-1}(\sin (-\frac{\pi }{6})-\cos (-\frac{\pi }{6}))\right]-\left[\sqrt{2}{\sin }^{-1}(\sin \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right))\right]$

$=\left[\sqrt{2}{\sin }^{-1}(-\sin \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right))\right]-\left[\sqrt{2}{\sin }^{-1}(\sin \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right))\right]$

$=\left[\sqrt{2}{\sin }^{-1}(-\frac{1}{2}-\frac{\sqrt{3}}{2})\right]-\left[\sqrt{2}{\sin }^{-1}(\frac{1}{2}-\frac{\sqrt{3}}{2})\right]$

$=\left[\sqrt{2}{\sin }^{-1}\left(\frac{-1-\sqrt{3}}{2}\right)\right]-\left[\sqrt{2}{\sin }^{-1}\left(\frac{1-\sqrt{3}}{2}\right)\right]$

$=\sqrt{2}\left[{\sin }^{-1}\left(\frac{-1-\sqrt{3}}{2}\right)\right]-\left[{\sin }^{-1}\left(\frac{1-\sqrt{3}}{2}\right)\right]$

Hence, this is the answer.