Friday, August 30, 2019

calculus - How to evaluate the following integral intpi/20sinxcosxln(sinx)ln(cosx),dx?



How to evaluate the following integral
π/20sinxcosxln(sinx)ln(cosx)dx
It seems that it evaluates to14π248
Is this true? How would I prove it?


Answer



Find this




I=π20sinxcosxln(cosx)ln(sinx)dx



Solution



Since
sin(2x)=2sin(x)cos(x)



then
I=18π20ln(sin2x)ln(cos2x)sin(2x)dx



Let cos(2x)=y, and since
cos(2x)=2cos2x1=12sin2x
we get
I=11611ln(1y2)ln(1+y2)dy



Let 1y2=z, then we have
I=1810lnzln(1z)dz=18n=11n10znlnzdz=18n=11n(n+1)2=18n=1(1n1n+1)18n=11(n+1)2=14π248


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