Wednesday, July 18, 2018

integration - Transformation to logleft(fraca2cosxright)




Question: How doesT=π2π2dxlog(a24cos2x)a24cos2x(a2cosx)=4π20dxlog(a2cosx)





I was evaluating an integral, namelyI=a0dxlogxaxx2And I'm having trouble seeing how to get from the left-hand side to the right-hand side. Doesn't the denominator and numerator cancel to leave you withT=π2π2dxlog(a24cos2x)=2π2π2dxlog(a2cosx)But I don't see how that can be substituted to get the second expression of (1). What kind of substitution should be made to make the transformation from where I left off to (1)?


Answer



First, cos(x)0 and is even for x[π/2,π/2]. Hence, cos2(x)=cos(x). If a>0, then a2=a. Therefore,



log(a24cos2(x))(a2cos(x))a24cos2(x)=log((acos(x)2)2)(a2cos(x))a2cos(x)=2log(acos(x)2)



Finally, exploiting the evenness of the cosine function, we find



π/2π/2log(a24cos2(x))(a2cos(x))a24cos2(x)dx=π/2π/22log(acos(x)2)dx=4π/20log(acos(x)2)dx



as expected!


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