Evaluate the integral using the indicated trigonometric substitution.

Question

Quizplus · Accepted Answer

The integral can be evaluated using the trigonometric substitution $x=3\sec	heta$.
Then, $dx=3\sec	heta	an	heta\ d	heta$ and $\sqrt{x^2-9}=3	an	heta$.
Substituting these into the integral, we get:
$$\int\frac{11}{\sqrt{x^2-9}}\ dx=\int\frac{11}{3	an	heta}\cdot 3\sec	heta	an	heta\ d	heta=\int 11\sec	heta\ d	heta$$
Using the identity $\sec^2	heta-1=	an^2	heta$, we get:
$$\int 11\sec	heta\ d	heta=11\ln(\sec	heta+	an	heta)+C$$
Substituting back $x=3\sec	heta$ and simplifying, we get the final answer:
$$\int\frac{11}{\sqrt{x^2-9}}\ dx=11\ln\left(\frac{x+\sqrt{x^2-9}}{3}ight)+C$$
Therefore, the best choice is D.

Evaluate the Integral Using the Indicated Trigonometric Substitution