Madilyn Fitzgerald

Answered

2022-01-28

Evaluation of ${\int}_{0}^{\frac{\pi}{2}}\frac{1}{{(a{\mathrm{cos}}^{2}\left(x\right)+b{\mathrm{sin}}^{2}\left(x\right))}^{n}}dx$

$n=1,2,3,\dots$

I thought about using$u=\mathrm{tan}x\text{}\text{or}\text{}u=\frac{\pi}{2}-x$ but did not work.

I thought about using

Answer & Explanation

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:

$\frac{\partial {I}_{n}}{\partial a}+\frac{\partial {I}_{n}}{\partial b}=-n{I}_{n+1}$

This recursion can be re-written alternatively as:

${I}_{n}=-\frac{1}{n-1}(\frac{\partial {I}_{n-1}}{\partial a}+\frac{\partial {I}_{n-1}}{\partial b}),{\textstyle \phantom{\rule{1em}{0ex}}}n=2,3,\dots$

and notice that$I}_{1$ can be evaluated rather easily using $u=\mathrm{tan}\left(x\right)$ to get $I}_{1}=\frac{\pi}{2\sqrt{ab}$

This recursion can be re-written alternatively as:

and notice that

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