Question

Evaluate the definite integral: $\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\frac{sinx+cosx}{\sqrt{sin2x}}dx$

Answer 1

$Let\space I=\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\frac{sinx+cosx}{\sqrt{sin2x}}dx$

⇒I=$$\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\space$$\frac{(sinx+cosx)}{\sqrt{-(-sin2x)}}dx

⇒I=$$\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\space$$\frac{sinx+cosx}{\sqrt{-(-1+1-2sinxcosx)}}dx

⇒I=$$\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\space$$\frac{(sinx+cosx)}{\sqrt{1-(sin^{2}x+cos^{2}x-2sinxcosx)}}dx

⇒I=$$\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\space $$\frac{(sinx+cosx)dx}{\sqrt{1-(sinx-cosx)^{2}}}

$Let(sinx-cosx)=t⇒(sinx+cosx)dx=dt$

$When x=\frac{\pi}{6},t=(\frac{1-\sqrt{3}}{2})and when x=\frac{\pi}{3},t=(\frac{\sqrt{3}-1}{2})$

$I=\int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^{2}}}$

⇒$$I=\int_{-({\frac{\sqrt{3}-1}{2})}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^{2}}}

$As \frac{1}{\sqrt{1-(-t)^{2}}}=\frac{1}{\sqrt{1-t^{2}}},therefore,\frac{1}{\sqrt{1-t^{2}}} is \space an\space even \space function.$

It is known that if $f(x)$ is an even function,then$\int^{a}_{-a}ƒ(x)dx=2\int^{a}_{0}ƒ(x)dx$

$⇒I=2\int^{\frac{\sqrt{3}-1}{2}}_{0} \frac{dt}{\sqrt{1-t^{2}}}$

$=[2sin^{-1}t]^{\frac{√3-1}{2}}_{0}$

$=2sin^{-1}(\frac{\sqrt{3}-1}{2})$