A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3Trig 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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3Trig 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" :)

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3Take 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?

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

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

agentx5
 2 years ago
Best 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 ;)

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

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

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

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

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

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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3SOHCAHTOA: 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?

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3We're going to use this to replace theta after you do that (now simple) integral above

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

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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.3That should be the answer I think ^_^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.