## anonymous 3 years ago Integrate 1/sinx and 1/cosx

1. 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?

2. anonymous

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

3. anonymous

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

4. anonymous

You are missing the point of the question, @zordoloom

5. anonymous

Seconded.

6. anonymous

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

7. anonymous

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

8. calculusfunctions

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

9. 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...

10. anonymous

Probably, but not the mnemonic

11. anonymous

No, looking on wikipedia I don't.

12. 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 ?

13. 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)

14. calculusfunctions

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

15. calculusfunctions

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

16. calculusfunctions

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

17. anonymous

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

18. anonymous

Oh dear, that was a stupid mistake, yes.

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

Could you explain why the LIATE 'rule' works?

23. anonymous

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

24. anonymous

$\int\limits \tan(x)\cos(x)x dx$

25. anonymous

Is the part of interest

26. anonymous

What exactly did I do incorrectly?

27. anonymous

What was incorrect?

28. 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.

29. anonymous

Where I wrote uv the bounds were implicit.

30. anonymous

It is the reverse of the product rule.

31. calculusfunctions

NO it is not!

32. anonymous

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

33. anonymous

@calculusfunctions , yes it actually is.

34. anonymous

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

35. 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."

36. anonymous

Carry on.

37. 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.

38. anonymous

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

39. 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!