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

Question

Quizplus · Accepted Answer

Substituting $x = 6\sec\theta$, we get $\sqrt{x^2 - 36} = 6\tan\theta$ and $dx = 6\sec\theta\tan\theta d\theta$. The integral becomes $\int 6\tan^2\theta d\theta$, which simplifies to $6\theta - 6\tan\theta + C$. Reverting back to $x$, we get $\sqrt{x^2 - 36} - 6\sec^{-1}\left(\frac{x}{6}\right) + C$.

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.

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