Evaluation of ∫0π21(acos2(x)+bsin2(x))ndx n=1,2,3,… I thought about using u=tanx or u=π2−x but did
I thought about using but did not work.
Answer & Explanation
2022-01-29Added 11 answers
Hint:Use Feynman’s Trick: differentiate the integral with respect to the parameters a and b, and it can be shown that:
This recursion can be re-written alternatively as:
and notice that can be evaluated rather easily using to get
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