Find the integral using an appropriate trigonometric substitution.

Question

Quizplus · Accepted Answer

Notice that the denominator $\sqrt{81-x^2}$ suggests that a trigonometric substitution $x=9\sin	heta$ would be convenient. Then we have $dx=9\cos	heta\ d	heta$ and

\begin{align*}
\sqrt{81-x^2}&=\sqrt{81-81\sin^2	heta} \
&= 9\cos	heta.
\end{align*}

Substituting $x=9\sin	heta$ and $dx=9\cos	heta\ d	heta$ gives:

\begin{align*}
\int \frac{9\cos	heta}{9\cos^2	heta}\ d	heta &= \int \frac{1}{\cos	heta}\ d	heta\
&= \ln|\sec	heta+	an	heta|+C.
\end{align*}

Finally, substituting back $	heta=\arcsin\frac{x}{9}$ gives:

$$\int \frac{1}{\sqrt{81-x^2}}\ dx = \ln\left|\frac{x}{\sqrt{81-x^2}}+\frac{\sqrt{81-x^2}}{9}ight|+C.$$

Therefore, the correct answer is B.

Find the Integral Using an Appropriate Trigonometric Substitution