anonymous
  • anonymous
Integrate 1/sinx and 1/cosx
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Which integrals method would you use IF YOU DID NOT KNOW WHAT DIFFERENTIATED TO GET IT (i.e. the answer) IN THE FIRST PLACE?
anonymous
  • anonymous
Do you know what 1/sinx is? and what 1/cosx is?
anonymous
  • anonymous
since 1/sinx is the same as cscx, You should look up in your book what the integration of cscx is.

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anonymous
  • anonymous
You are missing the point of the question, @zordoloom
anonymous
  • anonymous
Seconded.
anonymous
  • anonymous
I was contemplating an integration-by-parts idea after doing a substitution.
anonymous
  • anonymous
Assume I know everything that you're doing when solving, I'll say otherwise, but this problem seems very basic, just requiring insight.
calculusfunctions
  • calculusfunctions
OK! so you know that\[\int\limits_{}^{}udv =uv -\int\limits_{}^{}vdu\]Now do you know how to apply it your particular question?
anonymous
  • anonymous
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...
anonymous
  • anonymous
Probably, but not the mnemonic
anonymous
  • anonymous
No, looking on wikipedia I don't.
anonymous
  • anonymous
if i might give my opinion : multiply by sin(x) / sin(x) then change sin^2(x) in the denominator to 1-cos^2(x) .. can you integrate now ?
anonymous
  • anonymous
@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 1-cos^2(x)
calculusfunctions
  • calculusfunctions
@henpen are you there? I'm waiting for a response. Whenever you're ready.
calculusfunctions
  • calculusfunctions
@Coolsector I thought you meant multiply the given integrand by sin x.
calculusfunctions
  • calculusfunctions
No, it's not v. I said it's between u and dv.
anonymous
  • anonymous
Sorry, I always think by parts as\[\int\limits u\frac{dv}{dx}dx=uv-\int\limits \frac{du}{dx}vdx\]
anonymous
  • anonymous
Oh dear, that was a stupid mistake, yes.
anonymous
  • anonymous
Sorry, I've frozen up- I don't know
anonymous
  • anonymous
You're going to be integrating that, so not particularly.
anonymous
  • anonymous
Or am I being too self serving, choosing the easiest option (technically) ?
anonymous
  • anonymous
Could you explain why the LIATE 'rule' works?
anonymous
  • anonymous
\[\int\limits \csc(x)dx=\int\limits u \frac{dv}{dx}dx=uv-\int\limits \frac{du}{dx}vdx\] \[u=\csc(x), \frac{du}{dx}=\frac{-\cos(x)}{\sin^2(x)}\] \[\frac{dv}{dx}=1, v=x\]
anonymous
  • anonymous
\[\int\limits \tan(x)\cos(x)x dx\]
anonymous
  • anonymous
Is the part of interest
anonymous
  • anonymous
What exactly did I do incorrectly?
anonymous
  • anonymous
What was incorrect?
anonymous
  • anonymous
\[\frac{d(uv)}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u\] \[\frac{dv}{dx}u=\frac{d(uv)}{dx}-\frac{du}{dx}v\] Integrate both sides wrt x.
anonymous
  • anonymous
Where I wrote uv the bounds were implicit.
anonymous
  • anonymous
It is the reverse of the product rule.
calculusfunctions
  • calculusfunctions
NO it is not!
anonymous
  • anonymous
("It is the reverse of the product rule. " True Story)
anonymous
  • anonymous
@calculusfunctions , yes it actually is.
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Product_rule "The rule for integration by parts is derived from the product rule."
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Integration_by_parts "The rule can be derived in one line by simply integrating the product rule of differentiation."
anonymous
  • anonymous
Carry on.
anonymous
  • anonymous
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.
anonymous
  • anonymous
@calculusfunctions , this is getting nowhere: thank you for your time.
calculusfunctions
  • calculusfunctions
\[\int\limits_{}^{}\frac{ 1 }{ \sin x }dx\] \[=\int\limits_{}^{}\csc xdx\] \[=\ln \left| \csc x -\cot x \right|+C\]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!

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