`int_(0)^((pi)/(2))sin 2x tan^(-1)(sinx)dx`

Explanation

माना `I=int_(0)^(pi//2)sin 2x tan^(-1)(sinx)dx`
`=int_(0)^(pi//2)2sinxcos x tan^(-1)(sinx)dx`
माना ` sin x=t`
`rArr cos x dx = dt`
तथा `x=0" "rArr" "t=0,`
`" "x=(pi)/(2)" "rArr" "t=sin.(pi)/(2)=1`
`I=int_(0)^(1)2t tan^(-1)t dt = 2int_(0)^(1)underset("II")t.underset(" I ")(tan^(-1))tdt`
`=2[{tan^(-1)t(t^(2))/(2)}_(0)^(1)-int_(0)^(1)(1)/(1+t^(2)).(t^(2))/(2)dt]`
(खण्डशः समाकलन के प्रयोग से )
`=2((tan^(-1)1)/(2)-0)-int_(0)^(1)(t^(2))/(1+t^(2))dt`
`=2((pi)/(8))-int_(0)^(1)((1+t^(2))-1)/(1+t^(2))dt`
`=(pi)/(4)-int_(0)^(1)(1-(1)/(1_t^(2)))dt`
`=(pi)/(4)-[t-tan^(-1)t]_(0)^(1)`
`=(pi)/(4)-[1-tan^(-1)1-(0-0)]`
`=(pi)/(4)-1+(pi)/(4)=(pi)/(2)-1`