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Using Integration by parts

  1. Sep 1, 2004 #1
    [tex]\int 4x cos(2x)[/tex]

    using integration by parts...

    u=4x
    du= 4
    dv=cos(2x)
    v=cos(x)sin(x)

    using the formula uv - [tex]\int[/tex]v*du....

    4x*cos(x)sin(x) - [tex]\int[/tex]cos(x)sin(x)*4x

    hmm i cant seem to finish this problem, can someone help? and am i doing it correctly so far?
     
  2. jcsd
  3. Sep 1, 2004 #2

    Tide

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    Your expression for v is incorrect.
     
  4. Sep 1, 2004 #3

    arildno

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    DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!!
    Use the following:
    [tex]\int{uv'dx}=uv-\int{u'vdx}+C[/tex]
    Set:
    [tex]u(x)=4x, v'(x)=\cos(2x)\to{v(x)}=\frac{1}{2}\sin(2x)[/tex]
     
  5. Sep 2, 2004 #4

    ok so i plugged the parts into the formula
    [tex]\int{uv'dx}=uv-\int{u'vdx}+C[/tex]

    [tex]4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C[/tex]

    anti-deri. of sin is -cos that is why it's adding now...

    4x*1/2*sin(2x) + 4x*1/4*cos(2x)
    so... would that be the answer?
     
  6. Sep 2, 2004 #5
    Yes, it's the answer, try differentiating it.

    EDIT: Error.
     
    Last edited by a moderator: Sep 2, 2004
  7. Sep 2, 2004 #6
    the thing is, the answer is incorrect. i dont know what the correct answer is, but i know that isnt the answer. i submit my homework thru a online server and it checks your answer. what am i doing wrong?
     
  8. Sep 2, 2004 #7
    My deepest apologies. I must have been blind! The correct answer should be [tex]2xsin2x + cos2x[/tex] (written in Tex for clarity)
     
  9. Sep 2, 2004 #8
    thanks for the correct answer, but where did i go wrong?
     
  10. Sep 2, 2004 #9
    That x is not supposed to be there.
     
  11. Sep 3, 2004 #10
    [tex]4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C[/tex]

    hmm i thought i was suppose to take the anti-derivative of 4 also.... can you tell me why i dont need to take the anti-derivative of the "4" after the intergral?
     
  12. Sep 3, 2004 #11

    Math Is Hard

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    here are my thoughts..

    [tex] \int 4*1/2*sin(2x)dx[/tex]
    is equivalent to
    [tex] \int 2 *sin(2x)dx[/tex]
    is equivalent to
    [tex]2* \int sin(2x)dx[/tex]

    you can always pull out the constant factor
     
    Last edited: Sep 3, 2004
  13. Sep 5, 2004 #12

    HallsofIvy

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    What do you mean by "take the anti-derivative of 4 also"? Did you also think you should take the anti-derivative of the 1/2?
    [itex]\int{f(x)g(x)dx} [/itex] is NOT [itex]\int{f(x)dx}\int{g(x)dx}[/itex].

    In particular, if C is any constant (4, 1/2, whatever),
    [tex]\int{Cf(x)dx}= C\int{f(x)dx}[/itex].

    [tex]\int{4(1/2)sin(2x)dx}= \int{2sin(2x)dx}= 2\int{sin(2x)dx}= -cos(2x)+ C [/tex].
     
  14. Sep 6, 2004 #13

    that was very very very helpful, thank you sooo much. i always thought that [itex]\int{f(x)g(x)dx} [/itex] is equal to [itex]\int{f(x)dx}\int{g(x)dx}[/itex] .

    ok one more thing, so what is [itex]\int{f(x)g(x)dx} [/itex] then? would i just take the anti-derv. of f(x)?
     
  15. Sep 6, 2004 #14

    Tide

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    No. Generally, you would have to find the antiderivative of [itex]f(x)g(x)[/itex]. Knowing the antiderivative of one of them would permit you to use integration by parts.
     
  16. Sep 6, 2004 #15
    This is a little off-topic, but given a function, how do we know if it's anti-derivative exists?
     
  17. Sep 6, 2004 #16
    I think all continous functions have anti-derivatives, because given a real continous function f defined on say [a, b], if we define A by

    [tex]A(x) = \int^{x}_{a} f(t) dt[/tex]

    then A'(x) = f(x) (for all sensible values of x) (this is usually part of the proof of the fundamental theorem of (integral) calculus). So an anti-derivative exists (actually writing such a function in a closed form is a totally different matter though ;)).

    Someone correct me if I'm wrong...
     
    Last edited: Sep 6, 2004
  18. Sep 6, 2004 #17

    HallsofIvy

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    Every continuous function (in fact, every bounded function whose points of discontinuity form a set of measure 0) has an anti-derivative. Whether or not that anti-derivative is an "elementary" function depends on your definition of "elementary" function!

    No, of course, not! You can't just ignore g! The way you integrate a product IS "integration by parts".


    Remember the differentiation laws? IF (fg)'= f' g' THEN we could integrate both sides and get [itex]\int{f(x)g(x)dx}=\int{f(x)g(x)dx} [/itex].

    But that's NOT true. What is true is the "product rule": (fg)'= f'g+ fg'.
    That's the same as fg'= (fg)'- f'g. Integrating both sides,
    [itex]\intf(x)g'(x)dx= fg- \int g f' dx[/itex].

    Writing u= f, dv= g'dx, that is exactly [itex]\int udv= uv- \int vdu[/itex].
     
  19. Sep 6, 2004 #18

    i read your post many many times, and i kinda get what your saying, but i am also confused about some parts. im way better with examples, can you help me step by step and solve this problem:

    [tex]\int 4x cos(2x)[/tex] <-- this problem is the one i asked for from my first post, i already got the answer, but i dont know how Ethereal got the answer.

    let me show you how i tried to solve the problem.

    using [tex]\int{uv'dx}=uv-\int{u'vdx}+C[/tex]
    u(x) = 4x
    du(x) = 4dx
    dv(x) = cos(2x)
    v = 1/2sin(2x)

    plugged in what i have...

    [tex]4x*1/2sin(2x) \int 4*1/2sin(2x)[/tex]

    ok now i know that i need to solve [tex]\int 4*1/2sin(2x)[/tex]
    ok at this point, i am really stuck and confused. can someone help me step by step and explain to me what i really need to do? i know it's alot to ask for, but i really want to learn it.
     
    Last edited: Sep 6, 2004
  20. Sep 6, 2004 #19

    Math Is Hard

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    I think there's still some problems in your setup. The formula for int. by parts is:
    [itex]\int udv= uv- \int vdu[/itex]

    Let u = 4x
    dv = cos(2x)dx
    du = 4 dx
    v = 1/2 sin(2x)

    [itex]\int 4x cos(2x)dx = 4x (1/2 sin (2x)) - \int (1/2 sin (2x)) 4 dx[/itex]

    which becomes:

    [itex]\int 4x cos(2x)dx = 2x sin (2x) - \int 2 sin (2x) dx[/itex]

    which becomes

    [itex]\int 4x cos(2x)dx = 2x sin (2x) - 2\int sin (2x) dx[/itex]

    the last integral is pretty easy to solve with a substitution for 2x. Does that help at all?
     
  21. Sep 6, 2004 #20

    Math Is Hard

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    If you're stuck on the last part, show how you would integrate this:

    [itex]\int sin(2x)dx [/itex]

    that should pinpoint where the problem is.
     
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