Question: How doesT=−−π2∫π2dxlog(a24cos2x)√a24cos2x(a2cosx)=4π2∫0dxlog(a2cosx)
I was evaluating an integral, namelyI=a∫0dxlogx√ax−x2And 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=−−π2∫π2dxlog(a24cos2x)=−2−π2∫π2dxlog(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/2−π/2log(a24cos2(x))(a2cos(x))√a24cos2(x)dx=∫π/2−π/22log(acos(x)2)dx=4∫π/20log(acos(x)2)dx
as expected!
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