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agentx5Best ResponseYou've already chosen the best response.3
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
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
First while I go to get this food, 81 = 9^2 and 16 = 4^2
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
Trig sub. For this form with the 1/(x*sqrt(b^2*x^2a^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" :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
\[\int\limits \frac{1}{x\cdot (9^2x^24^2)^{1/2}}\cdot dx\] \[\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^24^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
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 tangentsquared now?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
lol ... the fastest way ... do opposite.
 one year ago

BeatlesBest ResponseYou've already chosen the best response.0
can we use the theorem of inverse trigo func. _
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
\[\int\limits \frac{1}{\sec\theta \cdot (9^2(\sec\theta)^24^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\] @experimentX actually faster way is to use the integral tables ;)
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
Err I mean wait that's 4/9 sec theta
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1342807447734:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1342807585448:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
well ... note the pattern .. rest is just some scrupulous manipulation.
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
\[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot (9^2(\frac{4}{9}\sec\theta)^24^2)^{1/2}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot \sqrt{16\tan^2\theta}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\frac{1}{9} \int\limits\limits \frac{9}{4} d\theta\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
still i recommend the trig substitution agentx5 did
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
Things cancel out, a lot. I get that on my scratchpad
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
SOHCAHTOA: 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?
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
We're going to use this to replace theta after you do that (now simple) integral above
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
well ... i call that more intuitive and geometrical approach.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1342808370968:dw the answer would be dw:1342808423489:dw sorry ... i forgot chain rule.
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
\[\huge \frac{1}{9} \int\limits \frac{9}{4} d\theta = \frac{1}{\cancel{9}} \left[ \frac{\cancel{9}\theta}{4} \right] = \frac{\sec^{1}(\frac{9 x}{4})}{4}+C\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.3
That should be the answer I think ^_^
 one year ago
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