Evaluation of ∫0π21(acos2(x)+bsin2(x))ndx n=1,2,3,… I thought about using u=tan⁡x or u=π2−x but did

Madilyn Fitzgerald

Madilyn Fitzgerald



Evaluation of 0π21(acos2(x)+bsin2(x))ndx
I thought about using u=tanx  or  u=π2x but did not work.

Answer & Explanation

Micah May

Micah May


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 I1 can be evaluated rather easily using u=tan(x) to get I1=π2ab

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