$\text { Find the indefinite integral by making the substitution } x = 6 \sec \theta \text {. }$ $$\int \frac { 1 } { \sqrt { x ^ { 2 } - 36 } } d x$$
A) $\ln \left| \sqrt { x ^ { 2 } + 36 } \right| + x ^ { 2 } + C$
B) $\ln \left| \sqrt { x ^ { 2 } - 36 } \right| + C$
C) $\ln \left| x \sqrt { x ^ { 2 } - 36 } \right| + C$
D) $\ln \left| \sqrt { x ^ { 2 } - 36 } + x \right| + C$
E) $\ln \left| x \sqrt { x ^ { 2 } - 36 } \right| + x + C$

Question

Quizplus · Accepted Answer

First, we make the substitution $x=6\sec	heta$. Therefore, $dx = 6\sec	heta	an	heta\,d	heta$ and $\sec^2	heta = \frac{x^2}{36} + 1$. Substituting these into the integral and simplifying, we get

\begin{align*}
\int\frac{1}{\sqrt{x^2-36}}\,dx &= \int\frac{1}{\sqrt{36\sec^2	heta-36}}\cdot 6\sec	heta	an	heta\,d	heta\
&=\int\frac{6\sec	heta	an	heta}{6\sqrt{\sec^2	heta-1}}\,d	heta\
&=\int	an	heta\,d	heta\
&=-\ln|\cos	heta| + C.
\end{align*}

Now we need to convert back to $x$ variables. Using the identity $\cos^2	heta=1-\sin^2	heta$, we can find that $\cos	heta = \frac{6}{\sqrt{x^2-36}}$. Therefore, substituting back into the integral, we have

\begin{align*}
\int\frac{1}{\sqrt{x^2-36}}\,dx &= -\ln\left|\frac{6}{\sqrt{x^2-36}}ight| + C\
&= \ln\left|\sqrt{x^2-36}-xight|+C.
\end{align*}

Thus, the answer is D.

Find the indefinite integral by making the substitution x=6sec⁡θ. \text { Find the indefinite integral by making the substitution } x = 6 \sec \theta \text {. } Find the indefinite integral by making the substitution x=6secθ.

$\text { Find the indefinite integral by making the substitution } x = 6 \sec \theta \text {. }$