# Evaluation of $\int_0^{\pi/2} \frac{1}{\left( a\cos^2(x)+b\sin^2(x) \right)^n} \, dx$

I would like to evaluate (using elementary methods if possible) : (for $a>0,\ b>0$)

$$I_n=\int_0^{\pi/2} \frac{1}{( a\cos^2(x)+b\sin^2(x))^n} \, dx,\quad \ n=1,2,3,\ldots$$ I thought about using $u=\tan(x)$ or $u=\frac{\pi}{2}-x$ but did not work. wolfram alpha evaluates the indefinite integral but not definite integral???

• you can use Feynman’s Trick – user547564 Aug 8 '18 at 21:05
• what is Feynman’s Trick??? – user579627 Aug 8 '18 at 21:32

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: $${{I}_{n}}=-\frac{1}{n-1}\left( \frac{\partial {{I}_{n-1}}}{\partial a}+\frac{\partial {{I}_{n-1}}}{\partial b} \right),\quad n=2,3,...$$ and notice that ${{I}_{1}}$ can be evaluated rather easily using $u=\tan \left( x \right)$ to get ${{I}_{1}}=\frac{\pi }{2\sqrt{ab}}$.

• Is it possible to use Feynman's Trick here $$\int_0^\frac{\pi}{2} \frac{\cos(A \cos(t))}{a^2 \sin^2 t + b^2 \cos^2 t}dt?$$ – Dinesh Shankar Aug 16 '18 at 20:36
• I am not sure, let me try – user547564 Aug 18 '18 at 7:12
• Thank you. If you can, let me know, so I'll post this as a new question. – Dinesh Shankar Aug 18 '18 at 11:53
• I will post a question about the first integral. – user547564 Aug 18 '18 at 14:36
• If you are interested. Here is a discussion about this second integral. – Dinesh Shankar Aug 18 '18 at 19:54

You may just use the residue theorem. $a$ and $b$ are interchangeable via $x\mapsto\frac{\pi}{2}-x$, so it is safe to assume $c\stackrel{\text{def}}{=}\frac{b}{a}\in(0,1)$ (since the case $a=b$ is trivial) and study $$I_n(a,b) = \frac{1}{a^n}\int_{0}^{\pi/2}\frac{d\theta}{(\cos^2\theta+c\sin^2\theta)^n}\stackrel{\theta\mapsto\arctan u}{=}\frac{1}{a^n}\int_{0}^{+\infty}\frac{du}{(1+u^2)\left(\frac{1+cu^2}{1+u^2}\right)^n}$$ which is $$\frac{1}{2a^n}\int_{\mathbb{R}}\frac{(1+u^2)^{n-1}}{(1+cu^2)^{n}}\,du =\frac{\pi i}{a^n}\operatorname*{Res}_{u=i/\sqrt{c}}\frac{(1+u^2)^{n-1}}{(1+cu^2)^n}.$$ $u=\frac{i}{\sqrt{c}}$ is clearly a pole of order $n$ for $\frac{(1+u^2)^{n-1}}{(1+cu^2)^n}$, hence the RHS equals $$\frac{\pi i}{(n-1)! a^n c^n}\lim_{u\to \frac{i}{\sqrt{c}}} \frac{d^{n-1}}{du^{n-1}}\frac{(1+u^2)^{n-1}}{\left(u+\frac{i}{\sqrt{c}}\right)^n}.$$

This is not an answer but it is too long for a comment.

For the antiderivative, Wolfram Alpha (and other CAS) return, for $$J_n(x)=(n-1)(a-b) I_n(x)$$s the messy expression $$2^{n-1} \csc (2 x) \sqrt{\frac{(a-b) \sin ^2(x)}{a}} \sqrt{\frac{(b-a) \cos ^2(x)}{b}}$$ $$((a-b) \cos (2 x)+a+b)^{1-n}$$ $$F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b+(a-b) \cos (2 x)}{2 b},\frac{a+b+(a-b) \cos (2 x)}{2 a}\right)$$ where appears

The problem is that both $J_n\left(\frac{\pi }{2}\right)$ and $J_n\left(0\right)$ result in indeterminate forms and that the limits need to be worked.

These would be $$J_n\left(\frac{\pi }{2}\right)=-\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}}\, b^{1-n}\, \Gamma (2-n)}{2 \, \Gamma \left(\frac{3}{2}-n\right)}\, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{b}{a}\right)$$ $$J_n\left(0\right)=\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}} \,a^{1-n}\, \Gamma (2-n)}{2\, \Gamma \left(\frac{3}{2}-n\right)}\, \, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{a}{b}\right)$$

Have fun !

• +1 thanks this is helpful – user579627 Aug 9 '18 at 16:29
• @user579627. You are welcome ! I wanted to explain why, having the antiderivative, may not correspond to the end of the work for the integral. Good example. Cheers. – Claude Leibovici Aug 9 '18 at 18:23