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How do you find:

$$\int_0^\pi\frac{dx}{(5+3\cos x)^3}$$

and

$$\int_0^\pi\frac{\sin^2x }{(5+3\cos x)^3}dx$$

by differentiating under the integral sign? I know you can get the result by using $\tan\left(\frac{x}{2}\right)$ substitution but I need to solve it via Feynman's Trick.

I am given the general result that:

$$\int_0^\pi\frac{dx}{a+b\cos x}= \frac{\pi}{\sqrt{a^2-b^2}}$$

Stefano
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Huwbaccaa
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  • $$f''(x) = 3\int_0^\pi\frac{\cos x}{(5+3\cos x)^2}\mathrm dx + 18\int_0^\pi\frac{\sin^2x}{(5+3\cos x)^2}\mathrm dx$$ Notice that the second integral here is the second in your post, and the first doesn't look that bad with the substitution $u = \cos(x)$. That's half the puzzle – Brevan Ellefsen Mar 07 '17 at 23:27

2 Answers2

4

$f(a) = \int_0^\pi\frac{dx}{(a+b\cos x)} = \frac {\pi}{\sqrt {a^2 - b^2}}\\ \frac {df}{da} = \int_0^\pi-(a+b\cos x)^{-2} dx = -a\pi(a^2 - b^2)^{-\frac 32}\\ \frac {d^2f}{da^2} = \int_0^\pi 2(a+b\cos x)^{-3}dx = \pi(-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})\\ \int_0^\pi (a+b\cos x)^{-3}dx = \frac \pi2 (-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})$

$\\$

$g(b) = \int_0^\pi (a+b\cos x)^{-1} dx = \pi(a^2 - b^2)^{-\frac 12}\\ g'(b) = \int_0^\pi -(\cos x)(a+b\cos x)^{-2} dx = b\pi(a^2 - b^2)^{-\frac 32}\\ \frac 12 g''(b) = \int_0^\pi (\cos^2 x)(a+b\cos x)^{-3} dx = \frac \pi2 ((a^2 - b^2)^{-\frac 32}+3b^2(a^2 - b^2)^{-\frac 52})\\ \int_0^\pi (\sin^2 x)(a+b\cos x)^{-3} dx = \frac 12(f''(a) - g''(b))$

Doug M
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2

Let $I(a,b)=\int_0^\pi \frac{1}{a+b\cos(x)}\,dx=\frac{\pi}{\sqrt{a^2-b^2}}$.

Then, we see that

$$\begin{align} \frac{\partial^2I(a,b)}{\partial b^2}&=\int_0^\pi \frac{2\cos^2(x)}{(a+b\cos(x))^3},dx\\\\ &=\color{blue}{2\int_0^\pi\frac{1}{(a+b\cos(x))^3}\,dx}-2\int_0^\pi \frac{\sin^2(x)}{(a+b\cos(x))^3}\,dx\\\\ &=\color{blue}{\frac{\partial^2I(a,b)}{\partial a^2}}-2\int_0^\pi \frac{\sin^2(x)}{(a+b\cos(x))^3}\,dx\\\\ \end{align}$$

Mark Viola
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