evaluate $int\frac{sqrt{x^{2}-49}}{x^{3}}dx$.

Answer:

$-\frac{\sqrt{x^{2}-49}}{2x^{2}}+\frac{1}{14}\sec^{-1}\left(\frac{x}{7}\right)+C$

Explanation:

Step1: Trigonometric substitution

Let $x = 7\sec\theta$, then $dx=7\sec\theta\tan\theta d\theta$. $\sqrt{x^{2}-49}=\sqrt{49\sec^{2}\theta - 49}=7\tan\theta$ and $x^{3}=343\sec^{3}\theta$. The integral becomes $\int\frac{7\tan\theta}{343\sec^{3}\theta}\cdot7\sec\theta\tan\theta d\theta=\frac{1}{7}\int\frac{\tan^{2}\theta}{\sec^{2}\theta}d\theta$.

Step2: Use trigonometric identity

Since $\tan^{2}\theta=\sec^{2}\theta - 1$, the integral is $\frac{1}{7}\int\frac{\sec^{2}\theta - 1}{\sec^{2}\theta}d\theta=\frac{1}{7}\int(1 - \cos^{2}\theta)d\theta$.

Step3: Integrate term - by - term

$\frac{1}{7}\int(1 - \cos^{2}\theta)d\theta=\frac{1}{7}\int\left(1-\frac{1 + \cos(2\theta)}{2}\right)d\theta=\frac{1}{7}\int\left(\frac{1}{2}-\frac{\cos(2\theta)}{2}\right)d\theta$. Integrating gives $\frac{1}{7}\left(\frac{1}{2}\theta-\frac{1}{4}\sin(2\theta)\right)+C$.

Step4: Use double - angle formula and back - substitution

$\sin(2\theta)=2\sin\theta\cos\theta$. Since $x = 7\sec\theta$, $\cos\theta=\frac{7}{x}$ and $\sin\theta=\frac{\sqrt{x^{2}-49}}{x}$. $\theta=\sec^{-1}\left(\frac{x}{7}\right)$. The integral is $\frac{1}{14}\sec^{-1}\left(\frac{x}{7}\right)-\frac{\sqrt{x^{2}-49}}{2x^{2}}+C$.