Question

Evaluate the definite integral ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

Answer 1

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$
$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2x)}}dx$
$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x+\cos x}{\sqrt{-(-1+1-2\sin x\cos x)}}dx$
$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x+\cos x)}{\sqrt{1-({\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x)}}dx$
$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x+\cos x)dx}{\sqrt{1-(\sin x-\cos x{)}^2}}$
Let $(\sin x-\cos x)=t\Rightarrow (\sin x+\cos x)dx=dt$
when $x=\frac{\pi }{6},t=\left(\frac{1-\sqrt{3}}{2}\right)$ and when $x=\frac{\pi }{3},t=\left(\frac{\sqrt{3}-1}{2}\right)$
$I={\int }_{\frac{1-\sqrt{3}}{3}}^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}}$
$\Rightarrow I={\int }_{-\left(\frac{\sqrt{3}-1}{2}\right)}^{\frac{\sqrt{3}-1}{1}}\frac{dt}{\sqrt{1-t^2}}$
As $\frac{1}{\sqrt{1-(-t{)}^2}}=\frac{1}{\sqrt{1-t^2}}$, therefore, $\frac{1}{\sqrt{1-t^2}}$ is an even function.
It is known that if f(x) is an even function, then ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$
$\Rightarrow I=2{\int }_0^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}}$
$=[2{\sin }^{-1}t{]}_0^{\frac{\sqrt{3}-1}{2}}$
$=2{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$