Question

Solve:
${\int }_0^{\pi /2}\sin 2x.{\tan }^{-1}\left(\sin x\right)dx$

Answer 1

${\int }_0^{\pi /2}sin2x.ta{n}^{-1}\left(sinx\right)dx=I.$

$I={\int }_0^{\pi /2}2sinxcosxta{n}^{-1}\left(sinx\right)dx$

Let $sinx=t\Rightarrow cosxdx=dt$

when $x=0,t=0$

when $x=\frac{\pi }{2},t=1$

so $I={\int }_0^{\pi /2}2sinxcosxta{n}^{-1}\left(t\right).\frac{dt}{cosx}$

$={\int }_0^1.2tcosxta{n}^{-1}\left(t\right).\frac{dt}{cosx}$

$={\int }_0^{1}2tta{n}^{-1}t.dt$

$=2{\int }_0^{1}tta{n}^{-1}t.dt$

$=2(ta{n}^{-1}t\int tdt-\int (\frac{d\left(ta{n}^{-1}t\right)}{dt}\int tdt\left)dt\right)$

$=2\left(ta{n}^{-1}t\right(\frac{t^2}{2})-\int \frac{1}{1+t^2}\times \frac{t^2}{2}.dt)$

$=2(\frac{t^2}{2}ta{n}^{-1}t-\frac{1}{2}\int \frac{t^2}{2}dt)$

$=t^{2}ta{n}^{-1}t-\int \frac{t^2}{1+t^2}dt.$ __ (1)

$\int \frac{t^2}{1+t^2}.dt=\int \frac{t^2+1-1}{t^2+1}.dt$

$=\int \frac{t^2+1}{t^2+1}-\frac{1}{t^2+1}.dt$

$=\int 1-\frac{1}{t^2+1}.dt.$

$=\int dt-\int \frac{dt}{t^2+1}$

$=t-ta{n}^{-1}t$

sub in (1)

$=t^{2}ta{n}^{-1}t-(t-ta{n}^{-1}t).$

$=t^{2}ta{n}^{-1}t-t+ta{n}^{-1}\left(t\right)$

$=F\left(x\right).$

$2{\int }_0^{1}tta{n}^{-1}t.dt$

$=F\left(1\right)-F\left(0\right)$

$=(1\times ta{n}^{-1}(1)-1+ta{n}^{-1}(1\left)\right)$

$(-0+ta{n}^{-1}(0\left)\right)$

$=\frac{\pi }{4}-1+\frac{\pi }{4}-(0-0+0)$

$=\frac{\pi }{2}-1-0$

$=\frac{\pi }{2}-1$