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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

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I wonder if many people know Calc II here...
\(x=4\sec\theta \implies dx=4\tan\theta \sec\theta d\theta \), substituting that we get: \[\int \frac{4\tan\theta \sec\theta}{16\sec^2\theta\sqrt{16\sec^2\theta-16}}d\theta=\int \frac{4\tan\theta}{16\sec\theta\cdot 4\tan\theta}=\frac{1}{16}\int \cos\theta=\cdots \]
Don't forget to substitute back for \(x\).

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Other answers:

\[dx=4sec( \theta) tan( \theta) d \theta \] \[\int\limits_{}^{}\frac{ 4 \sec(\theta) \tan(\theta) d \theta}{4^2 \sec^2(\theta) \sqrt{4 ^2 \sec^2(\theta)-16}} \] \[\frac{4}{4^2}\int\limits_{}^{}\frac{\tan(\theta) d \theta}{\sqrt{16} \sec(\theta) \sqrt{\sec^2(\theta)-1}}\] \[ \frac{1}{4} \cdot \frac{1}{\sqrt{16}} \int\limits_{}^{} \frac{\tan(\theta) d \theta}{\sec(\theta) \sqrt{\tan^2(\theta)}} \] \[\text{ Assume } \tan(\theta)>0 =>|\tan(\theta)|=\tan(\theta)\] 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 \[\frac{1}{16} \frac{\sqrt{x^2-16}}{x}+C\]
Right sqrt{x^2-16}.
Wow, thanks guys, that really helps a lot!
So @steelseries, remember the tricks :) to increase your probability of getting answers. OK?
Yeah, will do, thanks Captain!
What are those tricks? We want to know them too :-P

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