Question

Evaluate: `int_0^x (xtan x)/(sec x + tan x) dx`

Answer 1

`I = int_0^pi (x tan x)/(sec x + tan x) dx`      ......(1)

using `int_0^a f(x)dx = int_0^a f(a-x)dx`

`I = int_0^pi ((pi-x) tanpi -x)/(sec(pi-x)+ tan (pi -x))`

`I = int_0^pi ((pi - x) tan x) /(sec x + tan x )dx`                .....(2 ) `{

`{(sec(pi -x)= -secx),(tan (pi-x)=- tan x):}}`

Adding (1) and (2) we get
`therefore 2I = pi int_0^pi (tanx)/(secx + tan x)dx`

`2I = pi int_0^pi  (secx - tan x)dx)`

`2I = pi int_0^pi  (secx - tan x-tan^2x)dx)`

`2I = pi int_0^pi  (secx - tan x-tan^2x+1)dx)`

`2I = pi(secx - tanx + x)_0^pi`

`2I = pi int_0^pi  (sec pi - tan pi+pi) - (sec0 - tan 0+0)`

`I = pi/2 (-1 -0 + pi -(1 - 0 + 0) )`

`I = pi/2 (-1+ pi -1)`

`I = pi/2 (pi - 2)= pi (pi/2 - 1)`