Question

${\int }_0^{\frac{\pi }{2}}sin2xta{n}^{-1}\left(sinx\right)dx$

Answer 1

Let I = ${\int }_0^{\frac{\pi }{2}}sin2xta{n}^{-1}\left(sinx\right)dx$
${\int }_0^{\frac{\pi }{2}}sinxcosxta{n}^{-1}\left(sinx\right)dx[\because sin2x=2sinxcos]Putsinx=t\Rightarrow cosxdxdt\Rightarrow dx=\frac{dt}{cosx}when,x=0\Rightarrow t=0andwhenx=\frac{\pi }{2}\Rightarrow t=sin\frac{\pi }{2}=1\therefore I={\int }_0^{1}2tcosxta{n}^{-1}t\frac{dt}{cosx}=2{\int }_0^1\underset{II}{t}.\underset{I}{ta{n}^{-1}}tdt=2\left(\right[ta{n}^{-1}t\frac{t^2}{2}{]}_0^1-{\int }_0^1\frac{1}{1+t^2}.\frac{t^2}{2}dt)$ (Using intergration by parts)
$=2[\frac{ta{n}^{-1}1}{2}-0]-{\int }_0^1\frac{t^2}{1+t^2}dt=2\left(\frac{\pi }{8}\right)-{\int }_0^1\frac{(1+t^2)-1}{1+t^2}dt=\frac{\pi }{4}-{\int }_0^1(1-\frac{1}{1+t^2})dt=\frac{\pi }{4}-[t-ta{n}^{-1}t{]}_0^1=\frac{\pi }{4}-[1-ta{n}^{-1}1-(0-0)]=\frac{\pi }{4}-1+\frac{\pi }{4}=\frac{\pi }{2}-1$