evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-9}},x > 3 \nwhat…

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-9}},x > 3 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 3sin\theta)\nb. (x = 3sec\theta)\nc. (x = 3\tan\theta)

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-9}},x > 3 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 3sin\theta)\nb. (x = 3sec\theta)\nc. (x = 3\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 useful. Here $a = 3$ and we have $\sqrt{x^{2}-9}=\sqrt{x^{2}-3^{2}}$.

Step2: Analyze the options

If $x = 3\sec\theta$, then $x^{2}-9=9\sec^{2}\theta - 9=9(\sec^{2}\theta - 1)=9\tan^{2}\theta$ (using the identity $\sec^{2}\theta-1=\tan^{2}\theta$), and $\sqrt{x^{2}-9}=3|\tan\theta|$. Since $x>3$, when using $x = 3\sec\theta$, $\theta\in(0,\frac{\pi}{2})$ and $\tan\theta>0$, so $\sqrt{x^{2}-9}=3\tan\theta$. Also, $dx=3\sec\theta\tan\theta d\theta$.

Answer:

B. $x = 3\sec\theta$