Question:
Integrate : x+ sin x ÷1 +cos x with limits from 0 to π/2
Answer:
Let I = π/2∫0 (x + sinx)/(1 + cosx) dx
=> I = π/2∫0 {x + 2*sin(x/2)*cos(x/2)}/{2cos (x2 /2)}dx
=> I = π/2∫0 {x/{2cos2 x/2)}dx + π/2∫0 {2*sin(x/2)*cos(x/2)}/{2cos2 x/2)}dx
=> I = (1/2)* π/2∫0 x*sec2 x/2)}dx + π/2∫0 {sin(x/2)/cos(x/2)}dx
=> I = (1/2)* π/2∫0 x*sec2 x/2)}dx + π/2∫0 tan(x/2)dx
=> I = (1/2){[x*tan(x/2)/(1/2) 0]π/2 - π/2∫0 1*tan(x/2)/(1/2) * dx} + π/2∫0 tan(x/2)dx
=> I = {[x*tan(x/2) 0]π/2 - π/2∫0 tan(x/2)dx} + π/2∫0 tan(x/2)dx
=> I = [x*tan(x/2) 0]π/2
=> I = (π/2)*tan(π/4)
=> I = π/2
So, π/2∫0 (x + sinx)/(1 + cosx) dx = π/2
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