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agentx5 Group TitleBest 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
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
First while I go to get this food, 81 = 9^2 and 16 = 4^2
 2 years ago

agentx5 Group TitleBest 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" :)
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits \frac{1}{x\cdot (9^2x^24^2)^{1/2}}\cdot dx\] \[\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^24^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]
 2 years ago

agentx5 Group TitleBest 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?
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
Still there m8? :D
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
lol ... the fastest way ... do opposite.
 2 years ago

Beatles Group TitleBest ResponseYou've already chosen the best response.0
can we use the theorem of inverse trigo func. _
 2 years ago

agentx5 Group TitleBest 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 ;)
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
Err I mean wait that's 4/9 sec theta
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
Let me fix that...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1342807447734:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1342807585448:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
well ... note the pattern .. rest is just some scrupulous manipulation.
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot (9^2(\frac{4}{9}\sec\theta)^24^2)^{1/2}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot \sqrt{16\tan^2\theta}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\frac{1}{9} \int\limits\limits \frac{9}{4} d\theta\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
still i recommend the trig substitution agentx5 did
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
Things cancel out, a lot. I get that on my scratchpad
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
dw:1342807846986:dw
 2 years ago

agentx5 Group TitleBest 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?
 2 years ago

agentx5 Group TitleBest 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
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
well ... i call that more intuitive and geometrical approach.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1342808370968:dw the answer would be dw:1342808423489:dw sorry ... i forgot chain rule.
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
\[\huge \frac{1}{9} \int\limits \frac{9}{4} d\theta = \frac{1}{\cancel{9}} \left[ \frac{\cancel{9}\theta}{4} \right] = \frac{\sec^{1}(\frac{9 x}{4})}{4}+C\]
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.3
That should be the answer I think ^_^
 2 years ago
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