Improper integral $\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx$

Question

I can't figure out how to solve (say whether it converges or diverges) the following improper integral: $$ \int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx $$ I've tried to simplify this and got: $$ \int \limits_{2}^{4}\frac{(4 + x)^{5/2}}{(4 - x)^{1/2}(5 - x)^3} $$ But I don't know what to do next. Could you please give me some hint

Answer 1

In the absence of any other suggestions, here is an outline of how you can evaluate this by fairly straightforward means:

Firstly, from where you left off, you can do the substitution $x=4\cos 2\theta$ and obtain

$$1024\int_0^{\frac{\pi}{6}}\frac{\cos^6\theta}{(5-4\cos2\theta)^3}d\theta$$

Then a further substitution of $t=\tan\theta$ leads to $$1024\int_0^{\frac{1}{\sqrt{3}}}\frac{1}{(1+t^2)(1+9t^2)^3}dt$$

The integrand is then expressed in partial fractions, so we end up having to evaluate $$\int_0^{\frac{1}{\sqrt{3}}}\frac{18}{1+9t^2}-\frac{144}{(1+9t^2)^2}+\frac{1152}{(1+9t^2)^3}-\frac{2}{1+t^2}dt$$

And indeed the final answer is $36\sqrt{3}+\frac{125\pi}{3}$

Answer 2

An alternate substitution is $x = 4 \, \sin(t)$ which leads to, with some difficulty, \begin{align} \int \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = \frac{81}{2} \, \frac{x-6}{(x-5)^2} \, \sqrt{16 - x^2} - 63 \, \tan^{-1}\left(\frac{16 - 5x}{3 \, \sqrt{16 - x^2}}\right) - \sin^{-1}\left(\frac{x}{4}\right) + c_{0}. \end{align} Evaluating the integral at the limits (2,4) then provides the integrals value: \begin{align} \int_{2}^{4} \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = 36 \, \sqrt{3} + \frac{125 \, \pi}{3}. \end{align}