Question

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

Answer 1

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{sinx+cosx}{\sqrt{sin2x}}dx$
We know that $(sinx-cosx{)}^2=si{n}^{2}x+co{s}^{2}x-2$ sin x cos x
$\Rightarrow (sinx-cosx{)}^2=1-sin2x$
$\Rightarrow sin2x=1-(sinx-cosx{)}^2$
$\text{Put sin x - cos x = 1}$
$\Rightarrow (cosx+sinx)dx=dt\Rightarrow dx=\frac{dt}{cosx+sinx}\text{for limit when x =}\frac{\pi }{6}\Rightarrow t=sin\frac{\pi }{6}-cos\frac{\pi }{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}When,x=\frac{\pi }{3},t=sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}\therefore I={\int }_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{sinx+cosx}{\sqrt{1-t^2}}\frac{dt}{cosx+sinx}={\int }_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}}$
(From here we can solve it in two ways)
Ist method
$I=\int \frac{\frac{\sqrt{3}-1}{2}}{\frac{1-\sqrt{3}}{2}}\frac{1}{\sqrt{1-t^2}}dt=\left[si{n}^{-1}t\right]\frac{\frac{\sqrt{3}-1}{2}}{\frac{1-\sqrt{3}}{2}}=[si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)-si{n}^{-1}\{-\left(\frac{\sqrt{3}-1}{2}\right)\}]=[si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)+si{n}^{-1}\left(\frac{\sqrt{3-1}}{2}\right)][\because si{n}^{-1}(-\theta )=-si{n}^{-1}\theta ]=2si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$
IInd method
${\int }_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{1}{\sqrt{1-t^2}}dt={\int }_{(-\frac{\sqrt{3}-1}{2})}^{\frac{\sqrt{3}-1}{2}}\frac{1}{\sqrt{t-t^2}}dtAs\frac{1}{\sqrt{1-(t{)}^2}}=\frac{1}{\sqrt{1-t^2}},\text{therefore}\frac{1}{\sqrt{1-t^2}}\text{is an even function. and we know that if f(x) is an even function, then}{\int }_{-a}^{a}f\left(X\right)dx=2{}^{}{\int }_0^{a}f\left(X\right)dx\therefore I=2{\int }_0^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}}=[2si{n}^{-1}t{]}_0^{\frac{\sqrt{3}-1}{2}}=2si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$