Question

Integrate 1/sinx and 1/cosx

Answer 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?

Answer 2

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

Answer 3

since 1/sinx is the same as cscx, You should look up in your book what the integration of cscx is.

Answer 4

You are missing the point of the question, @zordoloom

Answer 5

Seconded.

Answer 6

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

Answer 7

Assume I know everything that you're doing when solving, I'll say otherwise, but this problem seems very basic, just requiring insight.

Answer 8

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

Answer 9

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...

Answer 10

Probably, but not the mnemonic

Answer 11

No, looking on wikipedia I don't.

Answer 12

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 ?

Answer 13

@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)

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Oh dear, that was a stupid mistake, yes.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Could you explain why the LIATE 'rule' works?

Answer 23

\[\int\limits \csc(x)dx=\int\limits u \frac{dv}{dx}dx=uv-\int\limits \frac{du}{dx}vdx\]
\[u=\csc(x), \frac{du}{dx}=\frac{-\cos(x)}{\sin^2(x)}\]
\[\frac{dv}{dx}=1, v=x\]

Answer 24

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

Answer 25

Is the part of interest

Answer 26

What exactly did I do incorrectly?

Answer 27

What was incorrect?

Answer 28

\[\frac{d(uv)}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u\]
\[\frac{dv}{dx}u=\frac{d(uv)}{dx}-\frac{du}{dx}v\]
Integrate both sides wrt x.

Answer 29

Where I wrote uv the bounds were implicit.

Answer 30

It is the reverse of the product rule.

Answer 31

NO it is not!

Answer 32

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

Answer 33

@calculusfunctions , yes it actually is.

Answer 34

http://en.wikipedia.org/wiki/Product_rule

"The rule for integration by parts is derived from the product rule."

Answer 35

http://en.wikipedia.org/wiki/Integration_by_parts

"The rule can be derived in one line by simply integrating the product rule of differentiation."

Answer 36

Carry on.

Answer 37

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.

Answer 38

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

Answer 39

\[\int\limits_{}^{}\frac{ 1 }{ \sin x }dx\]
\[=\int\limits_{}^{}\csc xdx\]
\[=\ln \left| \csc x -\cot x \right|+C\]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!