Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

henpen Group TitleBest ResponseYou've already chosen the best response.1
Which integrals method would you use IF YOU DID NOT KNOW WHAT DIFFERENTIATED TO GET IT (i.e. the answer) IN THE FIRST PLACE?
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Do you know what 1/sinx is? and what 1/cosx is?
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
since 1/sinx is the same as cscx, You should look up in your book what the integration of cscx is.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
You are missing the point of the question, @zordoloom
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
I was contemplating an integrationbyparts idea after doing a substitution.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Assume I know everything that you're doing when solving, I'll say otherwise, but this problem seems very basic, just requiring insight.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
OK! so you know that\[\int\limits_{}^{}udv =uv \int\limits_{}^{}vdu\]Now do you know how to apply it your particular question?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Your only choices for v and u are 1/sinx (to take 1st example) and 1, so I don't see the efficacy of that method...
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Probably, but not the mnemonic
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
No, looking on wikipedia I don't.
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
if i might give my opinion : multiply by sin(x) / sin(x) then change sin^2(x) in the denominator to 1cos^2(x) .. can you integrate now ?
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
@calculusfunctions if i multiply by sin(x)/sin(x) i get sin(x) / sin^2(x) this is a common method of solving such integrals (multiplying by a form of 1) then you change the sin^2(x) into 1cos^2(x)
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
@henpen are you there? I'm waiting for a response. Whenever you're ready.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
@Coolsector I thought you meant multiply the given integrand by sin x.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
No, it's not v. I said it's between u and dv.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Sorry, I always think by parts as\[\int\limits u\frac{dv}{dx}dx=uv\int\limits \frac{du}{dx}vdx\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Oh dear, that was a stupid mistake, yes.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Sorry, I've frozen up I don't know
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
You're going to be integrating that, so not particularly.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Or am I being too self serving, choosing the easiest option (technically) ?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Could you explain why the LIATE 'rule' works?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits \csc(x)dx=\int\limits u \frac{dv}{dx}dx=uv\int\limits \frac{du}{dx}vdx\] \[u=\csc(x), \frac{du}{dx}=\frac{\cos(x)}{\sin^2(x)}\] \[\frac{dv}{dx}=1, v=x\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits \tan(x)\cos(x)x dx\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Is the part of interest
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
What exactly did I do incorrectly?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
What was incorrect?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d(uv)}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u\] \[\frac{dv}{dx}u=\frac{d(uv)}{dx}\frac{du}{dx}v\] Integrate both sides wrt x.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Where I wrote uv the bounds were implicit.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
It is the reverse of the product rule.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
NO it is not!
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
("It is the reverse of the product rule. " True Story)
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
@calculusfunctions , yes it actually is.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Product_rule "The rule for integration by parts is derived from the product rule."
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Integration_by_parts "The rule can be derived in one line by simply integrating the product rule of differentiation."
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Carry on.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Given that it's more of a fundamental concept of calculus, you could argue that it's formed from the total differential of a function of 2 variables, but that's one step away from the product rule, so it makes little difference.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
@calculusfunctions , this is getting nowhere: thank you for your time.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ 1 }{ \sin x }dx\] \[=\int\limits_{}^{}\csc xdx\] \[=\ln \left \csc x \cot x \right+C\]There you go! I was merely trying to guide you to this conclusion. But instead of allowing me to help you opposed me every step of the way. GOOD LUCK!
 2 years ago
See more questions >>>
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.