Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Using Integration by parts

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

    using integration by parts...

    du= 4

    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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Use the following:
    [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]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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook