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What did i do wrong in my integration?

  1. Jan 25, 2007 #1

    dnt

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    1. The problem statement, all variables and given/known data

    (dont know how to make the integral sign so bare with me please)

    i will use | to signify the integral sign:

    here is my problem:

    |(sinx)(cos)dx

    2. Relevant equations

    n/a

    3. The attempt at a solution

    ok this is what i did:

    first i changed it to (1/2)sin2x because sin2x=2sinxcosx

    so now i have:

    |(1/2)sin2xdx

    then i said u = 2x and du=2dx

    so now i have:

    |(1/4)sinudu

    which is:

    (1/4)(-cos u) + C

    which is:

    (1/4)(-cos2x) + C

    where is my error?

    also, can someone give me a brief overview of the best strategy to figure out what type of integration to use? what do you look for to know when to use trig substituion, integration by parts, u-substitution, etc...

    ie, how do you know? thanks.
     
  2. jcsd
  3. Jan 25, 2007 #2
    let u = sin x. then du = cos x dx.

    Integration is just experience. You just need to practice.
     
  4. Jan 25, 2007 #3

    cristo

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    Alternatively, since you know the integral of sinx=-cosx, you should know that the integral of sin(nx)=-cos(nx)/n
     
  5. Jan 25, 2007 #4
    I had the exact same problem when I started integration, there really is no "trick" to know what type of integration to use. It all comes with ALOT of practice. Eventually you will just recognize things and you'll know what type of integration to use.
     
  6. Jan 25, 2007 #5
    There are general tips you can use.

    If you have a square root in the function of the form sqrt(a-x^2), sqrt(a+x^2), or sqrt(x^2-a), each has a different trig substitution.
    You use substitution to create a square trig function which will undo the root.

    By parts is best used if nothing is obvious (can't be more vague than that :P). Specifically if one part of the function is easily integratable (eg e^x, sin(x)).
    Use byparts when you see a product of somekind (hint: |f(x)dx can be considered a product of f(x) and dx)

    Partial fractions is merely an algebraic method for rewriting a function in an easier-to-integrate form. You largely only use this if you see a rational expression of one polynomial divided by another. (it may be particularly advantageous is the degree of the numerator > degree of denominator)

    substitution is useful if you see a function and it's derivative in the same function (eg: |ln(x)/x dx) But most times substitution can be done by inspection.
     
  7. Jan 26, 2007 #6

    Gib Z

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    Homework Helper

    What ChaoticLlama said is pretty good, but you'll learn those over practice as well. Only thing I have to add is Trig substitution is only good if you can do trigonometric integrals as well. I can always get it down to the trig integral, which is the aim, but i can't solve that one at the end..
     
  8. Jan 26, 2007 #7

    dnt

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    i should have added that i do know what the correct answer is and i knew how to get it. my question was in what step of my above work is wrong? i cant find it yet i know it comes out incorrect.
     
  9. Jan 26, 2007 #8
    I don't think it is, expand the double angle formula for cosine and I think you will find that your solution is equivalent to the other.
     
  10. Jan 26, 2007 #9

    HallsofIvy

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    What makes you think you have an error?

    Yes, the way most people would integrate [itex]\int sin(x)cos(x)dx[/itex] would be to let u= sin(x) so du= cos(x)dx and the integral becomes [itex]\int udu= \frac{1}{2}u^2+ C= \frac{1}{2}sin^2(x)+ C[/itex]

    But it is perfectly correct that sin(2x)= 2sin(x)cos(x) so that [itex]\int sin(x)cos(x)dx= \frac{1}{2}\int sin(2x)dx= \frac{1}{4}cos(2x)+ C'[/itex].

    Have you considered the possibility that
    [tex]\frac{1}{4}cos(2x)+ C'= \frac{1}{2}sin^2(x)+ C[/tex]
    possibly with different values for C and C'?

    Since you used sin(2x)= 2sin(x)cos(x) you might consider now using
    cos(2x)= cos2(x)- sin2= 1- 2sin2(x).
     
    Last edited: Jan 26, 2007
  11. Jan 26, 2007 #10
    You look at the answer and go :confused: You look at your answer and theirs again, and realise it's entirely equivalent because of the double angle formulas, I've been their :biggrin:

    :surprised :eek:
     
    Last edited: Jan 26, 2007
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