Saved Bookmarks
| 1. |
`int_(0)^(pi)(x tanx)/(secx+cosx)dx` isA. `(pi^(2))/4`B. `(pi^(2))/2`C. `(3pi^(2))/2`D. `(pi^(2))/3` |
|
Answer» Correct Answer - A Let `I=int_(0)^(pi)(x tan x)/(sec x +cosx)`……………1 `=int_(0)^(pi)((pi-x)tan(pi-x))/(sec(pi-x)+cos(pi-x))` `=int_(0)^(pi)((pi-x)tanx)/(secx+cosx) dx`…………….2 Adding 1 and 2 gives `2I=pi int_(0)^(pi) (tanx)/(secx+cosx)dx` `=pi int_(0)^(pi)((sinx)/(cosx))/(1/(cosx)+cosx)dx=pi int_(0)^(pi) (sinx)/(1+cos^(2)x) dx` Put `cosx=z`.Therefore `-sinxdx=dz` When `x=0, z=1`, when `x=pi,z=-1` `:.2I=pi int_(1)^(-1)(-dz)/(1+z^(2))=pi int_(-1)^(1)(dz)/(1+z^(2))` `=pi|tan^(-1)z|_(-1)^(1)` `=pi[tan^(-1)1-tan^(-1)(-1)]` `=pi((pi)/4+(pi)/4)=(2pi^(2))/4` or `I=(pi^(2))/4` |
|