The original problem is:
"Find the volume of the solid obtained by rotating about the x axis the region enclosed by the curves $y = \frac{9}{x^2 + 9},y=0,x=0,\,$and $x = 3$"
I set up the following integral $$81\pi\int_0^3\frac{1}{(x^2 + 9)^2}dx$$ using the cylinder method (I believe it's called like that) and when I calculated it using a computer I obtained the correct answer but I have been having difficulties in solving it manually. I tried the shell method as well and I didn't see it any easier to solve but I may be wrong of course.
$\endgroup$2 Answers
$\begingroup$We want to find $$\int \frac{dx}{(x^2+9)^2}.$$ Let $x=3u$. Apart from a constant factor, we end up with $$\int\frac{du}{(u^2+1)^2}.$$
Let $u=\tan t$ (we could have gone more directly, by letting $x=3\tan t$). Then $du=\sec^2 t\,dt$, and we end up with $$\int \frac{\sec^2 t}{\sec^4 t}\,dt.$$ This is the familiar integral $$\int \cos^2 t\,dt.$$ A common approach to this is to use the fact that $\cos 2t=2\cos^2 t-1$.
e can save time if we make the substitutions on the definite integral, that is, substitute for the endpoints.
Another way: Consider $\displaystyle\int \frac{dx}{x^2+9}$. Attack this by integration by parts, letting $du=dx$ and $v=\frac{1}{9+x^2}$. Then we can take $u=x$. Also, $dv=\frac{-2x\,dx}{(x^2+9)^2}$. Thus $$\int \frac{dx}{x^2+9}=\frac{x}{x^2+9}+\int\frac{2x^2\,dx}{(x^2+9)^2}.$$ But $2x^2=2x^2+18-18$. Thus $$\int\frac{dx}{x^2+9}=\frac{x}{x^2+9}+2\int\frac{dx}{x^2+9}-18\int\frac{dx}{(9+x^2)^2}.$$ We will be finished if we can find $\displaystyle\int\frac{dx}{x^2+9}$. The substitution $x=3w$ turns this into a familiar integral.
Remark: In the second approach, we obtained a reduction formula. A very similar trick expresses $\displaystyle\int \frac{dx}{(x^2+9)^{n+1}}$ in terms of $\displaystyle\int \frac{dx}{(x^2+9)^{n}}$
$\endgroup$ 2 $\begingroup$Although the integral you've got can be evaluated by some technical method, as @Andre noted completely; you can use The cylinder method instead. According to this method we consider the following integral: $$V=2\pi\int_{y=a}^{y=b}x|f(x)|dx$$ So we have here: $$\int_0^118\pi x\frac{1}{x^2+9}dx$$.
