InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Evaluate `int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`. |
|
Answer» Let `I=int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`…….`(i)` Then, `I=int_(0)^(pi)((pi-x)sin(pi-x))/(1+cos^(2)(pi-x))dx` or, `I=int_(0)^(pi)((pi-x)sinx)/((1+cos^(2)x))dx`……`(ii)` Adding `(i)` and `(ii)` , we get `2I=pi int_(0)^(pi)(sinx)/((1+cos^(2)x))dx=-piint_(1)^(-1)(dt)/((1+t^(2)))`, were `cosx=t` `=pi int_(-1)^(1)(dt)/((1+t^(2)))=pi[tan^(-1)t]_(-1)^(1)` `=pi[tan^(-1)-tan^(-1)]=pi[(pi)/(4)-(-(pi)/(4))]=(pi^(2))/(2)`. `:.I=(pi^(2))/(4)`. |
|