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Question Number 27502 by abdo imad last updated on 07/Jan/18
find∫0π2ln(1+xsin2t)sin2tdtwith−1<x<1.
Commented by abdo imad last updated on 09/Jan/18
letputf(x)=∫0π2ln(1+xsin2t)sin2tdtafterverifyingthatfisderivableon]−1,1[wehavef,(x)=∫0π2dt1+xsin2tbecauseof/xsin2t/<1f′(x)=∫0π2(∑n=0∝(−1)nxnsin2nt)dt=∑n=0∝(−1)nxn∫0π2sin2ntdt=∑n=0∝(−1)nWnxnwithWn=∫0π2sin2ntdtandthevalueofWnisknown(wallissintegral)f(x)=∑n=0∝(−1)nn+1Wnxn+1+λλ=f(0)=0⇒f(x)=∑n=0∝(−1)nn+1Wn.xn+1.
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