## Beatles 3 years ago $\int\limits \frac{dx}{x \sqrt{81x ^{2}-16}}$

1. agentx5

Trig Sub! I know how to do these now! :-D One moment, I'll show you steps... I need to get my lunch from the microwave oven :-3

2. agentx5

First while I go to get this food, 81 = 9^2 and 16 = 4^2

3. agentx5

Trig sub. For this form with the 1/(x*sqrt(b^2*x^2-a^a)).... let $$x = \large\frac{4}{9}\sec\theta$$ $$dx = \large\frac{4}{9}\tan\theta\sec\theta$$ @Beatles watch what happens when you put this in place for "x" and "dx" :-)

4. agentx5

$\int\limits \frac{1}{x\cdot (9^2x^2-4^2)^{1/2}}\cdot dx$ $\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)$

5. agentx5

Take the unit circle identity for sine and cosine, divide both side by cos^2 $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$ Can you see what happens here, @Beatles? Can you write the identity here in terms of tangent-squared now?

6. agentx5

Still there m8? :-D

7. experimentX

lol ... the fastest way ... do opposite.

8. Beatles

can we use the theorem of inverse trigo func. -_-

9. agentx5

$\int\limits \frac{1}{\sec\theta \cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)$ @experimentX actually faster way is to use the integral tables ;-)

10. agentx5

Err I mean wait that's 4/9 sec theta

11. agentx5

Let me fix that...

12. experimentX

|dw:1342807447734:dw|

13. experimentX

|dw:1342807585448:dw|

14. experimentX

well ... note the pattern .. rest is just some scrupulous manipulation.

15. agentx5

$\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot (9^2(\frac{4}{9}\sec\theta)^2-4^2)^{1/2}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)$ $\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot \sqrt{16\tan^2\theta}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)$ $\frac{1}{9} \int\limits\limits \frac{9}{4} d\theta$

16. experimentX

still i recommend the trig substitution agentx5 did

17. agentx5

Things cancel out, a lot. I get that on my scratchpad

18. agentx5

|dw:1342807846986:dw|

19. agentx5

SOH-CAH-TOA: SOH : $$\sin \theta = \large \frac{opposite}{hypotenuse}$$ CAH : $$\cos \theta = \large \frac{adjacent}{hypotenuse}$$ TOA : $$\tan \theta = \large \frac{opposite}{adjacent}$$ $$\huge \frac{1}{\cos\theta} = \sec\theta$$ Therefore: $$\huge \theta = sec^{^-1}(\frac{hyp}{adj})=sec^{-1}(\frac{9x}{4})$$ Agreed @TuringTest & @experimentX , and do you follow what I'm doing here so far @Beatles?

20. agentx5

We're going to use this to replace theta after you do that (now simple) integral above

21. experimentX

well ... i call that more intuitive and geometrical approach.

22. experimentX

|dw:1342808370968:dw| the answer would be |dw:1342808423489:dw| sorry ... i forgot chain rule.

23. agentx5

$\huge \frac{1}{9} \int\limits \frac{9}{4} d\theta = \frac{1}{\cancel{9}} \left[ \frac{\cancel{9}\theta}{4} \right] = \frac{\sec^{-1}(\frac{9 x}{4})}{4}+C$

24. agentx5

That should be the answer I think ^_^