Saturday, 24 June 2017

definite integrals - Integrating displaystyleintpi/20sin2xover1+sinxcosxdx



We need to evaluate π/20sin2x1+sinxcosxdx
and some solution to this starts as,




π/20sin2x1+sinxcosxdx=π/20{sin(π/2x)}21+sin(π/2x)cos(π/2x)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+bx) dx,




I=π20sin2x1+sinxcosxdx=π20cos2x1+sinxcosxdx



2I=π2011+sinxcosxdx=π20sec2x1+tan2x+tanxdx



Setting tanx=u



2I=0dt1+t+t2=40dt(2t+1)2+(3)2



Set 2t+1=3tanθ


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