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henpen

  • 2 years ago

Integrate 1/sinx and 1/cosx

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  1. henpen
    • 2 years ago
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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?

  2. zordoloom
    • 2 years ago
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    Do you know what 1/sinx is? and what 1/cosx is?

  3. zordoloom
    • 2 years ago
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    since 1/sinx is the same as cscx, You should look up in your book what the integration of cscx is.

  4. CliffSedge
    • 2 years ago
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    You are missing the point of the question, @zordoloom

  5. henpen
    • 2 years ago
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    Seconded.

  6. CliffSedge
    • 2 years ago
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    I was contemplating an integration-by-parts idea after doing a substitution.

  7. henpen
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Probably, but not the mnemonic

  11. henpen
    • 2 years ago
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    No, looking on wikipedia I don't.

  12. Coolsector
    • 2 years ago
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    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
    • 2 years ago
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    @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
    • 2 years ago
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    @henpen are you there? I'm waiting for a response. Whenever you're ready.

  15. calculusfunctions
    • 2 years ago
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    @Coolsector I thought you meant multiply the given integrand by sin x.

  16. calculusfunctions
    • 2 years ago
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    No, it's not v. I said it's between u and dv.

  17. henpen
    • 2 years ago
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    Sorry, I always think by parts as\[\int\limits u\frac{dv}{dx}dx=uv-\int\limits \frac{du}{dx}vdx\]

  18. henpen
    • 2 years ago
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    Oh dear, that was a stupid mistake, yes.

  19. henpen
    • 2 years ago
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    Sorry, I've frozen up- I don't know

  20. henpen
    • 2 years ago
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    You're going to be integrating that, so not particularly.

  21. henpen
    • 2 years ago
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    Or am I being too self serving, choosing the easiest option (technically) ?

  22. henpen
    • 2 years ago
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    Could you explain why the LIATE 'rule' works?

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

  24. henpen
    • 2 years ago
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    \[\int\limits \tan(x)\cos(x)x dx\]

  25. henpen
    • 2 years ago
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    Is the part of interest

  26. henpen
    • 2 years ago
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    What exactly did I do incorrectly?

  27. henpen
    • 2 years ago
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    What was incorrect?

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

  29. henpen
    • 2 years ago
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    Where I wrote uv the bounds were implicit.

  30. henpen
    • 2 years ago
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    It is the reverse of the product rule.

  31. calculusfunctions
    • 2 years ago
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    NO it is not!

  32. CliffSedge
    • 2 years ago
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    ("It is the reverse of the product rule. " True Story)

  33. CliffSedge
    • 2 years ago
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    @calculusfunctions , yes it actually is.

  34. CliffSedge
    • 2 years ago
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    http://en.wikipedia.org/wiki/Product_rule "The rule for integration by parts is derived from the product rule."

  35. CliffSedge
    • 2 years ago
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    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
    • 2 years ago
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    Carry on.

  37. henpen
    • 2 years ago
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    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
    • 2 years ago
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    @calculusfunctions , this is getting nowhere: thank you for your time.

  39. calculusfunctions
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ 1 }{ \sin x }dx\] \[=\int\limits_{}^{}\csc xdx\] \[=\ln \left| \csc x -\cot x \right|+C\]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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