## 2bornot2b 3 years ago 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

1. SteelSeries

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

2. Mr.Math

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

3. Mr.Math

Don't forget to substitute back for $$x$$.

4. myininaya

$dx=4sec( \theta) tan( \theta) d \theta$ $\int\limits_{}^{}\frac{ 4 \sec(\theta) \tan(\theta) d \theta}{4^2 \sec^2(\theta) \sqrt{4 ^2 \sec^2(\theta)-16}}$ $\frac{4}{4^2}\int\limits_{}^{}\frac{\tan(\theta) d \theta}{\sqrt{16} \sec(\theta) \sqrt{\sec^2(\theta)-1}}$ $\frac{1}{4} \cdot \frac{1}{\sqrt{16}} \int\limits_{}^{} \frac{\tan(\theta) d \theta}{\sec(\theta) \sqrt{\tan^2(\theta)}}$ $\text{ Assume } \tan(\theta)>0 =>|\tan(\theta)|=\tan(\theta)$ 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 $\frac{1}{16} \frac{\sqrt{x^2-16}}{x}+C$

5. SteelSeries

Woah

6. Mr.Math

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7. myininaya

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8. Mr.Math

Right sqrt{x^2-16}.

9. SteelSeries

Wow, thanks guys, that really helps a lot!

10. 2bornot2b

So @steelseries, remember the tricks :) to increase your probability of getting answers. OK?

11. SteelSeries

Yeah, will do, thanks Captain!

12. Mr.Math

What are those tricks? We want to know them too :-P

13. 2bornot2b

If so then here is the real story. http://openstudy.com/users/steelseries#/updates/4f2c1192e4b0e1bedb5d1d04

14. saifoo.khan