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Integrate 1/sinx and 1/cosx

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Which integrals method would you use IF YOU DID NOT KNOW WHAT DIFFERENTIATED TO GET IT (i.e. the answer) IN THE FIRST PLACE?
Do you know what 1/sinx is? and what 1/cosx is?
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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Other answers:

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