How to evaluate the following integral
∫π/20sinxcosxln(sinx)ln(cosx)dx
It seems that it evaluates to14−π248
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=18∫π20ln(sin2x)ln(cos2x)sin(2x)dx
Let cos(2x)=y, and since
cos(2x)=2cos2x−1=1−2sin2x
we get
I=116∫1−1ln(1−y2)ln(1+y2)dy
Let 1−y2=z, then we have
I=18∫10lnzln(1−z)dz=−18∞∑n=11n∫10znlnzdz=18∞∑n=11n(n+1)2=18∞∑n=1(1n−1n+1)−18∞∑n=11(n+1)2=14−π248
No comments:
Post a Comment