evaluate the integral.\n intsqrt{\frac{x^{2}-8}{x^{8}}}dx \n intsqrt{\frac{x^{2}-8}{x^{8}}}dx=square

Explanation:

Step1: Simplify the integrand

First, simplify $\sqrt{\frac{x^{2}-8}{x^{8}}}=\frac{\sqrt{x^{2}-8}}{x^{4}}$. Let $x = 2\sqrt{2}\sec\theta$, then $dx=2\sqrt{2}\sec\theta\tan\theta d\theta$. And $\sqrt{x^{2}-8}=\sqrt{8\sec^{2}\theta - 8}=2\sqrt{2}\tan\theta$, $x^{4}=64\sqrt{4}\sec^{4}\theta$.

Step2: Substitute into the integral

The integral $\int\frac{\sqrt{x^{2}-8}}{x^{4}}dx=\int\frac{2\sqrt{2}\tan\theta}{64\sqrt{4}\sec^{4}\theta}\cdot2\sqrt{2}\sec\theta\tan\theta d\theta=\frac{1}{8}\int\frac{\tan^{2}\theta}{\sec^{3}\theta}d\theta$. Since $\tan^{2}\theta=\sec^{2}\theta - 1$, we have $\frac{1}{8}\int\frac{\sec^{2}\theta - 1}{\sec^{3}\theta}d\theta=\frac{1}{8}\int(\cos\theta-\cos^{3}\theta)d\theta$.

Step3: Integrate term - by - term

We know that $\int\cos\theta d\theta=\sin\theta + C_1$ and $\int\cos^{3}\theta d\theta=\int\cos\theta(1 - \sin^{2}\theta)d\theta$. Let $u = \sin\theta$, then $du=\cos\theta d\theta$, and $\int\cos\theta(1 - \sin^{2}\theta)d\theta=\int(1 - u^{2})du=u-\frac{u^{3}}{3}+C_2=\sin\theta-\frac{\sin^{3}\theta}{3}+C_2$. So $\frac{1}{8}\int(\cos\theta-\cos^{3}\theta)d\theta=\frac{1}{8}(\sin\theta-\sin\theta+\frac{\sin^{3}\theta}{3})+C=\frac{1}{24}\sin^{3}\theta + C$.

Step4: Back - substitute

Since $x = 2\sqrt{2}\sec\theta$, then $\sec\theta=\frac{x}{2\sqrt{2}}$, $\cos\theta=\frac{2\sqrt{2}}{x}$, and $\sin\theta=\frac{\sqrt{x^{2}-8}}{x}$. So the integral $\int\sqrt{\frac{x^{2}-8}{x^{8}}}dx=-\frac{(x^{2}-8)^{\frac{3}{2}}}{24x^{3}}+C$.

Answer:

$-\frac{(x^{2}-8)^{\frac{3}{2}}}{24x^{3}}+C$