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