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

Answered

2022-01-28

Evaluation of 0π21(acos2(x)+bsin2(x))ndx
n=1,2,3,
I thought about using u=tanx  or  u=π2x but did not work.

Answer & Explanation

Micah May

Micah May

Expert

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:
Ina+Inb=nIn+1
This recursion can be re-written alternatively as:
In=1n1(In1a+In1b),n=2,3,
and notice that I1 can be evaluated rather easily using u=tan(x) to get I1=π2ab

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