Integrate 1/sinx and 1/cosx

- anonymous

Integrate 1/sinx and 1/cosx

- katieb

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- 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

Do you know what 1/sinx is? and what 1/cosx is?

- 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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## More answers

- anonymous

You are missing the point of the question, @zordoloom

- anonymous

Seconded.

- anonymous

I was contemplating an integration-by-parts idea after doing a substitution.

- 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

OK! so you know that\[\int\limits_{}^{}udv =uv -\int\limits_{}^{}vdu\]Now do you know how to apply it your particular question?

- 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

Probably, but not the mnemonic

- anonymous

No, looking on wikipedia I don't.

- 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

@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

@henpen are you there? I'm waiting for a response. Whenever you're ready.

- calculusfunctions

@Coolsector I thought you meant multiply the given integrand by sin x.

- calculusfunctions

No, it's not v. I said it's between u and dv.

- anonymous

Sorry, I always think by parts as\[\int\limits u\frac{dv}{dx}dx=uv-\int\limits \frac{du}{dx}vdx\]

- anonymous

Oh dear, that was a stupid mistake, yes.

- anonymous

Sorry, I've frozen up- I don't know

- anonymous

You're going to be integrating that, so not particularly.

- anonymous

Or am I being too self serving, choosing the easiest option (technically)
?

- anonymous

Could you explain why the LIATE 'rule' works?

- 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

\[\int\limits \tan(x)\cos(x)x dx\]

- anonymous

Is the part of interest

- anonymous

What exactly did I do incorrectly?

- anonymous

What was incorrect?

- 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

Where I wrote uv the bounds were implicit.

- anonymous

It is the reverse of the product rule.

- calculusfunctions

NO it is not!

- anonymous

("It is the reverse of the product rule. " True Story)

- anonymous

@calculusfunctions , yes it actually is.

- anonymous

http://en.wikipedia.org/wiki/Product_rule
"The rule for integration by parts is derived from the product rule."

- 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

Carry on.

- 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

@calculusfunctions , this is getting nowhere: thank you for your time.

- 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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