`int_(0)^(pi)(x tanx)/(secx+cosx)dx` is

Options

A. `(pi^(2))/4`
B. `(pi^(2))/2`
C. `(3pi^(2))/2`
D. `(pi^(2))/3`

Answer

A

Explanation

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`