evaluate the integral. (use c for the constant of integration.)\n$$ int \frac { sqrt { x ^ { 2 } - 36 } } {…

Explanation:

Step1: Use trigonometric substitution

Let (x = 6\sec\theta), then (dx=6\sec\theta\tan\theta d\theta). And (\sqrt{x^{2}-36}=\sqrt{36\sec^{2}\theta - 36}=6\tan\theta), (x^{4}=(6\sec\theta)^{4}). The integral (\int\frac{\sqrt{x^{2}-36}}{x^{4}}dx=\int\frac{6\tan\theta}{(6\sec\theta)^{4}}\cdot6\sec\theta\tan\theta d\theta). Simplify the integrand: [ \begin{align*} \int\frac{6\tan\theta}{(6\sec\theta)^{4}}\cdot6\sec\theta\tan\theta d\theta&=\frac{1}{216}\int\frac{\tan^{2}\theta}{\sec^{3}\theta}d\theta\ &=\frac{1}{216}\int\sin^{2}\theta\cos\theta d\theta \end{align*} ]

Step2: Use substitution (u = \sin\theta)

Let (u=\sin\theta), then (du=\cos\theta d\theta). The integral (\frac{1}{216}\int\sin^{2}\theta\cos\theta d\theta=\frac{1}{216}\int u^{2}du). Integrate (\int u^{2}du=\frac{u^{3}}{3}+C). So (\frac{1}{216}\int u^{2}du=\frac{u^{3}}{648}+C).

Step3: Back - substitute

Since (u = \sin\theta) and (x = 6\sec\theta) (i.e., (\cos\theta=\frac{6}{x}), (\sin\theta=\frac{\sqrt{x^{2}-36}}{x})). [ \begin{align*} \frac{u^{3}}{648}+C&=\frac{(\frac{\sqrt{x^{2}-36}}{x})^{3}}{648}+C\ &=\frac{(x^{2}-36)^{\frac{3}{2}}}{648x^{3}}+C \end{align*} ]

Answer:

(\frac{(x^{2}-36)^{\frac{3}{2}}}{648x^{3}}+C)