Friday, 9 May 2014

calculus - Putnam definite integral evaluation intpi/20fracxsinxcosxsin4x+cos4xdx




Evaluate π/20xsinxcosxsin4x+cos4xdx Source : Putnam




By the property a0f(x)dx=a0f(ax)dx:



=π/20(π/2x)sinxcosxsin4x+cos4xdx=π2π/20sinxcosxsin4x+cos4xdxπ/20xsinxcosxsin4x+cos4xdx




π/20xsinxcosxsin4x+cos4xdx=π4π/20sinxcosxsin4x+cos4xdx



Now I'm stuck. WolframAlpha says the indefinite integral of sinxcosxsin4x+cos4x evaluates nicely to 12arctan(cos(2x)).



I already factored sin4x+cos4x into 1(sin(2x)2)2, but I don't know how to continue.. I suggest a substitution u=sin(2x)2?



Could someone provide me a hint, or maybe an easier method I can refer to in the future?


Answer



1sin2(2x)2=1+cos2(2x)2, and sinxcosx=sin(2x)2sinxcosxcos4x+sin4xdx=12d(cos(2x))1+cos2(2x)dx=12arctan(cos(2x))+C



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