21. - / 1.07 points evaluate the integral. $int_{2}^{9}\frac{dx}{(x^{2}-1)^{3/2}}$

Explanation:

Step1: Use trigonometric substitution

Let $x = \sec\theta$, then $dx=\sec\theta\tan\theta d\theta$. When $x = 2$, $\sec\theta=2\Rightarrow\theta=\frac{\pi}{3}$; when $x = 9$, $\sec\theta = 9\Rightarrow\theta=\text{arcsec}(9)$. Also, $(x^{2}-1)^{\frac{3}{2}}=(\sec^{2}\theta - 1)^{\frac{3}{2}}=\tan^{3}\theta$.

Step2: Rewrite the integral

The integral $\int_{2}^{9}\frac{dx}{(x^{2}-1)^{\frac{3}{2}}}$ becomes $\int_{\frac{\pi}{3}}^{\text{arcsec}(9)}\frac{\sec\theta\tan\theta}{\tan^{3}\theta}d\theta=\int_{\frac{\pi}{3}}^{\text{arcsec}(9)}\frac{\sec\theta}{\tan^{2}\theta}d\theta=\int_{\frac{\pi}{3}}^{\text{arcsec}(9)}\frac{\cos\theta}{\sin^{2}\theta}d\theta$.

Step3: Use substitution for the new - integral

Let $u=\sin\theta$, then $du=\cos\theta d\theta$. When $\theta=\frac{\pi}{3}$, $u = \frac{\sqrt{3}}{2}$; when $\theta=\text{arcsec}(9)$, $u=\frac{\sqrt{80}}{9}$. The integral $\int_{\frac{\pi}{3}}^{\text{arcsec}(9)}\frac{\cos\theta}{\sin^{2}\theta}d\theta=\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{80}}{9}}\frac{du}{u^{2}}$.

Step4: Evaluate the integral

We know that $\int\frac{du}{u^{2}}=-\frac{1}{u}+C$. So $\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{80}}{9}}\frac{du}{u^{2}}=\left[-\frac{1}{u}\right]_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{80}}{9}}=-\frac{9}{\sqrt{80}}+\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}-\frac{9}{4\sqrt{5}}=\frac{8\sqrt{5}-27\sqrt{3}}{12\sqrt{15}}$.

Answer:

$\frac{8\sqrt{5}-27\sqrt{3}}{12\sqrt{15}}$