I'm lost on how to pull the anti-derivative of the expression under the radical. The full integral I'm evaluating is:
$$2\pi\int_{16}^{25}(9-{\sqrt{x}})^2{\sqrt{1+\left(\frac{9}{\sqrt{x}}\right)^2}}dx$$
I think I need to use u-substitution, however I'm not sure what my u should equal.
Let $\sqrt{x}=9\tan{t},$ where $t\in\left(0,\frac{\pi}{2}\right).$
Thus, $$\frac{1}{2\sqrt{x}}dx=\frac{9}{\cos^2t}dt$$ or $$dx=\frac{162\sin{t}}{\cos^3t}dt.$$
Id est, $$\int\left(9-\sqrt{x}\right)^2\sqrt{1+\left(\frac{9}{\sqrt{x}}\right)^2}dx=\int(9-9\tan{t})^2\sqrt{1+\cot^2t}\frac{162\sin{t}}{\cos^3t}dt=$$ $$=13122\int\frac{(1-2\sin{t}\cos{t})\sin{t}}{\cos^2t\sin{t}\cos^3t}dt=13122\left(\int\frac{1}{\cos^5t}dt-2\int\frac{\sin{t}}{\cos^4t}dt\right)=$$ $$=13122\left(\int\frac{1}{(1-\sin^2t)^3}d\sin{t}+2\int\frac{1}{\cos^4t}d\cos{t}\right)=...$$ Can you end it now?