Evaluate ∫ [sinx/(cosx + sinx) dx on the interval [0,π/2]?
I initially used a trick method using ∫ f(a-x)dx on [0,a] and I got π/4, but he wants me to also do it using another "known integration" method.
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∫ sin(x) dx / [(cos(x) + sin(x)]
multiply with (cosx - sinx ) / ( cosx - sinx)
∫sin(x) [cosx - sinx] dx / (cos^2(x) - sin^2(x)
∫[sin(x)cosx - sin^2(x)] dx / cos(2x)
∫{(1/2)[sin(2x)- (1/2)[(1 - cos(2x)]} dx / cos(2x)
=1/2 ∫tan(2x) dx - 1/2 ∫ sec(2x) dx + 1/2∫ dx
= (1/4) ln [sec(2x) ] - (1/4) ln [sec(2x) + tan(2x)] + (1/2) x +c
The substitution u=tan(x/2) will turn any rational function of sinx and cosx into the integral of a rational function of u. With this substitution
sinx = 2u/(1+u^2), cosx = (1-u^2)/(1+u^2), and dx = 2du/(1+u^2)
Your integral then is the integral of a rational function of u, that can be converted into partial fractions. The limits of integration wrt u will be 0 to 1.
For details, see the link below.
∫ [cosx/(cosx + sinx) dx