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anonymous
 4 years ago
Integrate 1/sinx and 1/cosx
anonymous
 4 years ago
Integrate 1/sinx and 1/cosx

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Which integrals method would you use IF YOU DID NOT KNOW WHAT DIFFERENTIATED TO GET IT (i.e. the answer) IN THE FIRST PLACE?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know what 1/sinx is? and what 1/cosx is?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since 1/sinx is the same as cscx, You should look up in your book what the integration of cscx is.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You are missing the point of the question, @zordoloom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was contemplating an integrationbyparts idea after doing a substitution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Assume I know everything that you're doing when solving, I'll say otherwise, but this problem seems very basic, just requiring insight.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your 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
 4 years ago
Best ResponseYou've already chosen the best response.0Probably, but not the mnemonic

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, looking on wikipedia I don't.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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 ?

anonymous
 4 years ago
Best 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)

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I always think by parts as\[\int\limits u\frac{dv}{dx}dx=uv\int\limits \frac{du}{dx}vdx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh dear, that was a stupid mistake, yes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I've frozen up I don't know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You're going to be integrating that, so not particularly.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or am I being too self serving, choosing the easiest option (technically) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could you explain why the LIATE 'rule' works?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \csc(x)dx=\int\limits u \frac{dv}{dx}dx=uv\int\limits \frac{du}{dx}vdx\] \[u=\csc(x), \frac{du}{dx}=\frac{\cos(x)}{\sin^2(x)}\] \[\frac{dv}{dx}=1, v=x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \tan(x)\cos(x)x dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is the part of interest

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What exactly did I do incorrectly?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d(uv)}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u\] \[\frac{dv}{dx}u=\frac{d(uv)}{dx}\frac{du}{dx}v\] Integrate both sides wrt x.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Where I wrote uv the bounds were implicit.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is the reverse of the product rule.

calculusfunctions
 4 years ago
Best ResponseYou've already chosen the best response.0NO it is not!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0("It is the reverse of the product rule. " True Story)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@calculusfunctions , yes it actually is.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Given 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
 4 years ago
Best ResponseYou've already chosen the best response.0@calculusfunctions , this is getting nowhere: thank you for your time.

calculusfunctions
 4 years ago
Best 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!
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