consider the following function.\n int \frac { x ^ { 2 } } { sqrt { x ^ { 2 } - 5 } } d x \n(a) determine an…

consider the following function.\n int \frac { x ^ { 2 } } { sqrt { x ^ { 2 } - 5 } } d x \n(a) determine an appropriate trigonometric substitution.\n \begin{array} { l } { \text { use } x = sqrt { 5 } sin ( \theta ) , \text { where } - \frac { pi } { 2 } leq \theta leq \frac { pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } \\ { \text { use } x = sqrt { 5 } \tan ( \theta ) , \text { where } - \frac { pi } { 2 } < \theta < \frac { pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } \\ { \text { use } x = sqrt { 5 } sec ( \theta ) , \text { where } 0 leq \theta < \frac { pi } { 2 } \text { or } pi leq \theta < \frac { 3 pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } end{array} \n(b) apply the substitution to transform the integral into a trigonometric integral. do not evaluate the integral.\n int \frac { x ^ { 2 } } { sqrt { x ^ { 2 } - 5 } } d x = int ( quad ) d \theta

consider the following function.\n int \frac { x ^ { 2 } } { sqrt { x ^ { 2 } - 5 } } d x \n(a) determine an appropriate trigonometric substitution.\n \begin{array} { l } { \text { use } x = sqrt { 5 } sin ( \theta ) , \text { where } - \frac { pi } { 2 } leq \theta leq \frac { pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } \\ { \text { use } x = sqrt { 5 } \tan ( \theta ) , \text { where } - \frac { pi } { 2 } < \theta < \frac { pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } \\ { \text { use } x = sqrt { 5 } sec ( \theta ) , \text { where } 0 leq \theta < \frac { pi } { 2 } \text { or } pi leq \theta < \frac { 3 pi } { 2 } , \text { since the integrand contains the expression } sqrt { x ^ { 2 } - 5 } } end{array} \n(b) apply the substitution to transform the integral into a trigonometric integral. do not evaluate the integral.\n int \frac { x ^ { 2 } } { sqrt { x ^ { 2 } - 5 } } d x = int ( quad ) d \theta

Answer

Explanation:

Step1: Substitute (x = \sqrt{5}\sec\theta)

Differentiate (x) with respect to (\theta): (dx=\sqrt{5}\sec\theta\tan\theta d\theta)

Step2: Substitute (x) and (dx) into the integral

  • (x^{2}=5\sec^{2}\theta)
  • (\sqrt{x^{2}-5}=\sqrt{5\sec^{2}\theta - 5}=\sqrt{5(\sec^{2}\theta - 1)}=\sqrt{5\tan^{2}\theta}=\sqrt{5}\vert\tan\theta\vert). Since (0\leq\theta<\frac{\pi}{2}) or (\pi\leq\theta<\frac{3\pi}{2}), (\tan\theta) has the same sign as (x) (in the domain of substitution), and (\vert\tan\theta\vert = \tan\theta)
  • The integral (\int\frac{x^{2}}{\sqrt{x^{2}-5}}dx=\int\frac{5\sec^{2}\theta}{\sqrt{5}\tan\theta}\cdot\sqrt{5}\sec\theta\tan\theta d\theta=\int5\sec^{3}\theta d\theta)

Answer:

(\int5\sec^{3}\theta d\theta)