evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-289}},x > 17…

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-289}},x > 17 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 17sec\theta)\nb. (x = 17\tan\theta)\nc. (x = 17sin\theta)

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-289}},x > 17 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 17sec\theta)\nb. (x = 17\tan\theta)\nc. (x = 17sin\theta)

Answer

Explanation:

Step1: Recall trig - substitution rules

For integrals of the form $\int\frac{dx}{\sqrt{x^{2}-a^{2}}}$ with $x > a$, the substitution $x=a\sec\theta$ is used. Here $a = 17$ since $x^{2}-289=x^{2}-17^{2}$.

Step2: Check the substitution

If $x = 17\sec\theta$, then $dx=17\sec\theta\tan\theta d\theta$ and $\sqrt{x^{2}-289}=\sqrt{289\sec^{2}\theta - 289}=17\tan\theta$ (since $\sec^{2}\theta-1=\tan^{2}\theta$ and for the domain $x>17$, $\tan\theta>0$).

Answer:

A. $x = 17\sec\theta$