anonymous
  • anonymous
\[\int\limits \frac{dx}{x \sqrt{81x ^{2}-16}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
First while I go to get this food, 81 = 9^2 and 16 = 4^2
anonymous
  • anonymous
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" :-)

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anonymous
  • anonymous
\[\int\limits \frac{1}{x\cdot (9^2x^2-4^2)^{1/2}}\cdot dx\] \[\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
Still there m8? :-D
experimentX
  • experimentX
lol ... the fastest way ... do opposite.
anonymous
  • anonymous
can we use the theorem of inverse trigo func. -_-
anonymous
  • anonymous
\[\int\limits \frac{1}{\sec\theta \cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\] @experimentX actually faster way is to use the integral tables ;-)
anonymous
  • anonymous
Err I mean wait that's 4/9 sec theta
anonymous
  • anonymous
Let me fix that...
experimentX
  • experimentX
|dw:1342807447734:dw|
experimentX
  • experimentX
|dw:1342807585448:dw|
experimentX
  • experimentX
well ... note the pattern .. rest is just some scrupulous manipulation.
anonymous
  • anonymous
\[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot (9^2(\frac{4}{9}\sec\theta)^2-4^2)^{1/2}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot \sqrt{16\tan^2\theta}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\frac{1}{9} \int\limits\limits \frac{9}{4} d\theta\]
experimentX
  • experimentX
still i recommend the trig substitution agentx5 did
anonymous
  • anonymous
Things cancel out, a lot. I get that on my scratchpad
anonymous
  • anonymous
|dw:1342807846986:dw|
anonymous
  • anonymous
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?
anonymous
  • anonymous
We're going to use this to replace theta after you do that (now simple) integral above
experimentX
  • experimentX
well ... i call that more intuitive and geometrical approach.
experimentX
  • experimentX
|dw:1342808370968:dw| the answer would be |dw:1342808423489:dw| sorry ... i forgot chain rule.
anonymous
  • anonymous
\[\huge \frac{1}{9} \int\limits \frac{9}{4} d\theta = \frac{1}{\cancel{9}} \left[ \frac{\cancel{9}\theta}{4} \right] = \frac{\sec^{-1}(\frac{9 x}{4})}{4}+C\]
anonymous
  • anonymous
That should be the answer I think ^_^

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