evaluate $int_{4}^{10}\frac{dx}{(9x^{2}-36)^{\frac{3}{2}}}$

Explanation:

Step1: Factor out constant

First, factor out 9 from the denominator: $\int_{4}^{10}\frac{dx}{(9(x^{2}-4))^{\frac{3}{2}}}=\frac{1}{27}\int_{4}^{10}\frac{dx}{(x^{2}-4)^{\frac{3}{2}}}$.

Step2: Trigonometric - substitution

Let $x = 2\sec\theta$, then $dx=2\sec\theta\tan\theta d\theta$. When $x = 4$, $\sec\theta = 2\Rightarrow\theta=\frac{\pi}{3}$; when $x = 10$, $\sec\theta = 5\Rightarrow\theta=\sec^{- 1}(5)$. Also, $x^{2}-4 = 4\sec^{2}\theta - 4=4\tan^{2}\theta$. So the integral becomes $\frac{1}{27}\int_{\frac{\pi}{3}}^{\sec^{-1}(5)}\frac{2\sec\theta\tan\theta d\theta}{(4\tan^{2}\theta)^{\frac{3}{2}}}=\frac{1}{27}\int_{\frac{\pi}{3}}^{\sec^{-1}(5)}\frac{2\sec\theta\tan\theta d\theta}{8\tan^{3}\theta}=\frac{1}{108}\int_{\frac{\pi}{3}}^{\sec^{-1}(5)}\frac{\sec\theta}{\tan^{2}\theta}d\theta$. Since $\frac{\sec\theta}{\tan^{2}\theta}=\frac{\cos\theta}{\sin^{2}\theta}$, let $u=\sin\theta$, then $du=\cos\theta d\theta$.

Step3: Integrate

The integral $\frac{1}{108}\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{24}}{5}}\frac{du}{u^{2}}=\frac{1}{108}\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{24}}{5}}u^{-2}du$. Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{1}{108}\left[-\frac{1}{u}\right]_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{24}}{5}}$.

Step4: Evaluate definite integral

$\frac{1}{108}\left(-\frac{5}{\sqrt{24}}+\frac{2}{\sqrt{3}}\right)=\frac{1}{108}\left(-\frac{5}{2\sqrt{6}}+\frac{2}{\sqrt{3}}\right)=\frac{1}{108}\left(-\frac{5\sqrt{6}}{12}+\frac{4\sqrt{6}}{6}\right)=\frac{1}{108}\left(-\frac{5\sqrt{6}}{12}+\frac{8\sqrt{6}}{12}\right)=\frac{\sqrt{6}}{432}$.

Answer:

$\frac{\sqrt{6}}{432}$