Question

Evaluate the integral \[\int\frac{1}{(x^2\sqrt{x^2-16)}}dx\] using the substitution \[x=4\sec\theta\]

This question is originally posted by SteelSeries

Answer 1

I wonder if many people know Calc II here...

Answer 2

\(x=4\sec\theta \implies dx=4\tan\theta \sec\theta d\theta \), substituting that we get:
\[\int \frac{4\tan\theta \sec\theta}{16\sec^2\theta\sqrt{16\sec^2\theta-16}}d\theta=\int \frac{4\tan\theta}{16\sec\theta\cdot 4\tan\theta}=\frac{1}{16}\int \cos\theta=\cdots \]

Answer 3

Don't forget to substitute back for \(x\).

Answer 4

\[dx=4sec( \theta) tan( \theta) d \theta \]
\[\int\limits_{}^{}\frac{ 4 \sec(\theta) \tan(\theta) d \theta}{4^2 \sec^2(\theta) \sqrt{4 ^2 \sec^2(\theta)-16}} \]
\[\frac{4}{4^2}\int\limits_{}^{}\frac{\tan(\theta) d \theta}{\sqrt{16} \sec(\theta) \sqrt{\sec^2(\theta)-1}}\]
\[ \frac{1}{4} \cdot \frac{1}{\sqrt{16}} \int\limits_{}^{} \frac{\tan(\theta) d \theta}{\sec(\theta) \sqrt{\tan^2(\theta)}} \]
\[\text{ Assume } \tan(\theta)>0 =>|\tan(\theta)|=\tan(\theta)\]
So we have
\[\frac{1}{16}\int\limits_{}^{}\cos(\theta) d \theta= \frac{1}{16}\sin(\theta)+C\]
Recall the sub:
\[x=4 \sec(\theta) => \sec(\theta)=\frac{x}{4} (=\frac{hyp}{adj})\]
So opposite=...
\[opp=\sqrt{x^2-4^2}=\sqrt{x^2-16}\]
So we have the answer is
\[\frac{1}{16} \frac{\sqrt{x^2-16}}{x}+C\]

Answer 5

Woah