evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13…

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 13sin\theta)\nb. (x = 13sec\theta)\nc. (x = 13\tan\theta)

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 13sin\theta)\nb. (x = 13sec\theta)\nc. (x = 13\tan\theta)

Answer

Explanation:

Step1: Recall trig - substitution rules

For $\sqrt{x^{2}-a^{2}}$ with $x > a$, the substitution $x = a\sec\theta$ is used. Here $a = 13$ and the integrand has $\sqrt{x^{2}-169}=\sqrt{x^{2}-13^{2}}$.

Step2: Analyze the substitution options

If $x = 13\sec\theta$, then $dx=13\sec\theta\tan\theta d\theta$ and $\sqrt{x^{2}-169}=\sqrt{169\sec^{2}\theta - 169}=13\tan\theta$ (since $\sec^{2}\theta-1 = \tan^{2}\theta$ and for the domain $x>13$, $\theta\in(0,\frac{\pi}{2})$ where $\tan\theta>0$).

Answer:

B. $x = 13\sec\theta$