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henpenBest 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?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Do you know what 1/sinx is? and what 1/cosx is?
 one year ago

zordoloomBest 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.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
You are missing the point of the question, @zordoloom
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
I was contemplating an integrationbyparts idea after doing a substitution.
 one year ago

henpenBest 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.
 one year ago

calculusfunctionsBest 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?
 one year ago

henpenBest 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...
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Probably, but not the mnemonic
 one year ago

henpenBest ResponseYou've already chosen the best response.1
No, looking on wikipedia I don't.
 one year ago

CoolsectorBest 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 ?
 one year ago

CoolsectorBest 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)
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
@henpen are you there? I'm waiting for a response. Whenever you're ready.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
@Coolsector I thought you meant multiply the given integrand by sin x.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
No, it's not v. I said it's between u and dv.
 one year ago

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

henpenBest ResponseYou've already chosen the best response.1
Oh dear, that was a stupid mistake, yes.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Sorry, I've frozen up I don't know
 one year ago

henpenBest ResponseYou've already chosen the best response.1
You're going to be integrating that, so not particularly.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Or am I being too self serving, choosing the easiest option (technically) ?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Could you explain why the LIATE 'rule' works?
 one year ago

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

henpenBest ResponseYou've already chosen the best response.1
\[\int\limits \tan(x)\cos(x)x dx\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Is the part of interest
 one year ago

henpenBest ResponseYou've already chosen the best response.1
What exactly did I do incorrectly?
 one year ago

henpenBest 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.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Where I wrote uv the bounds were implicit.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
It is the reverse of the product rule.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
NO it is not!
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
("It is the reverse of the product rule. " True Story)
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
@calculusfunctions , yes it actually is.
 one year ago

CliffSedgeBest 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."
 one year ago

CliffSedgeBest 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."
 one year ago

henpenBest 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.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
@calculusfunctions , this is getting nowhere: thank you for your time.
 one year ago

calculusfunctionsBest 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!
 one year ago
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