## henpen Group Title Integrate 1/sinx and 1/cosx one year ago one year ago

1. henpen Group Title

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. zordoloom Group Title

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

3. zordoloom Group Title

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

4. CliffSedge Group Title

You are missing the point of the question, @zordoloom

5. henpen Group Title

Seconded.

6. CliffSedge Group Title

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

7. henpen Group Title

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 Group Title

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

9. henpen Group Title

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. henpen Group Title

Probably, but not the mnemonic

11. henpen Group Title

No, looking on wikipedia I don't.

12. Coolsector Group Title

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. Coolsector Group Title

@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 Group Title

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

15. calculusfunctions Group Title

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

16. calculusfunctions Group Title

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

17. henpen Group Title

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

18. henpen Group Title

Oh dear, that was a stupid mistake, yes.

19. henpen Group Title

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

20. henpen Group Title

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

21. henpen Group Title

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

22. henpen Group Title

Could you explain why the LIATE 'rule' works?

23. henpen Group Title

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

24. henpen Group Title

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

25. henpen Group Title

Is the part of interest

26. henpen Group Title

What exactly did I do incorrectly?

27. henpen Group Title

What was incorrect?

28. henpen Group Title

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

29. henpen Group Title

Where I wrote uv the bounds were implicit.

30. henpen Group Title

It is the reverse of the product rule.

31. calculusfunctions Group Title

NO it is not!

32. CliffSedge Group Title

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

33. CliffSedge Group Title

@calculusfunctions , yes it actually is.

34. CliffSedge Group Title

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

35. CliffSedge Group Title

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. CliffSedge Group Title

Carry on.

37. henpen Group Title

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. henpen Group Title

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

39. calculusfunctions Group Title

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