Question

Evaluate the following : `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x)*dx`

Answer 1

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

= `int_0^(pi/4) (cos^2x - sin^2x)/(2cos^2x + 2 sin x cosx)*dx`

= `int_0^(pi/4) ((cosx - sinx)(cosx + sinx))/(2cosx(cosx + sinx))*dx`

= `int_0^(pi/4) (cosx - sinx)/(2cosx)*dx`

= `(1)/(2) int_0^(pi/4) [cosx/cosx - sinx/cosx]*dx`

= `(1)/(2) [int_0^(pi/4) 1*dx - int_0^(pi/4) tanx*dx]`

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

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

= `(1)/(2)[pi/4 - log sqrt(2) + log 1]`

= `(1)/(2)[pi/4 - log sqrt(2)]`.                    ...[∵ log 1 = 0]