We need to evaluate ∫π/20sin2x1+sinxcosxdx
and some solution to this starts as,
∫π/20sin2x1+sinxcosxdx=∫π/20{sin(π/2−x)}21+sin(π/2−x)cos(π/2−x)dx.
We fail to understand how this step has been carried out. Even variable substitution
does not seem to work.
Do you think that you could give us a hint?
Answer
Using ∫baf(x) dx=∫baf(a+b−x) dx,
I=∫π20sin2x1+sinxcosxdx=∫π20cos2x1+sinxcosxdx
2I=∫π2011+sinxcosxdx=∫π20sec2x1+tan2x+tanxdx
Setting tanx=u
2I=∫∞0dt1+t+t2=4∫∞0dt(2t+1)2+(√3)2
Set 2t+1=√3tanθ
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