InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`int_(0)^((pi)/(2))sin 2x tan^(-1)(sinx)dx` |
|
Answer» 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` |
|