Evaluate: `intsin2xtan^(-1)(sinx)\ dx`

Explanation

Given integral is,
`intsin2xtan^(-1)(sinx)\ dx`
`=int 2sinx cosx tan^(-1)(sinx)\ dx`
taking `sinx=t`
`=>cosxdx=dt`
putting these values ,given integral becomes,
`=2 int t tan^-1 (t) dt`
`=2[tan^-1 t]tdt-int (d(tan^-1t))/(dt)(int tdt)dt]`
`=2[t^2/2 tan^-1(t)-int(1/(1+t^2))(t^2/2)dt`
`=t^2 tan^-1(t)-int((t^2+1-1)/(1+t^2))dt`
`=t^2 tan^-1(t)-int 1-1/(1+t^2) dt`
`=t^2 tan^-1(t)-t+tan^-1(t)+c`
`=sin^2xtan^-1 (sinx)-sinx+tan^-1(sinx)+c`