Question

Evaluate: `int_0^(pi/4)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`

Answer 1

Let I = `int_0^(pi/4)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`

= `int_0^(pi/4)  (cos^2x - sin^2x)/(2cos^2x + 2sinx cosx)  "d"x`

= `int_0^(pi/4)  ((cosx + sinx)(cosx - sin x))/(2cos(cosx + sinx))  "d"x`

= `1/2 int_0^(pi/4)  (cosx - sinx)/cosx  "d"x`

= `1/2 int_0^(pi/4) (1 - tan x)  "d"x`

= `1/2 int_0^(pi/4)  "d"x - 1/2 int_0^(pi/4) tanx  "d"x`

= `1/2[x]_0^(pi/4) - 1/2[log|sec x|]_0^(pi/4)`

= `1/2(pi/4 - 0) - 1/2[log|sec  pi/4| - log|sec 0|]`

= `pi/8 - 1/2 (log sqrt(2) - log 1)`

= `pi/8 - 1/2 (log sqrt(2) - 0)`

∴ I = `1/2(pi/4 - log sqrt(2))`