antiderivatives, calculus

Discussion in 'Homework Help' started by learninmath, May 19, 2008.

  1. learninmath

    Thread Starter New Member

    May 19, 2008
    5
    0
    HELP PLEASE!

    i got the question; find, sin^4x dx in antiderivatives

    and got two possible answers: 1/2x+1/4cos4x+c
    and: 1/2sin^4x+c

    im so confused:confused:
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    do you mean the integral of sin(4x) dx ?
     
  3. learninmath

    Thread Starter New Member

    May 19, 2008
    5
    0
    yep sorry to miss that out
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    the answer is -(1/4)cos(4x)+C
     
  5. triggernum5

    Active Member

    May 4, 2008
    216
    0
    How did you approach the question to get those? Like mik, I'm assuming you want to find Integral[sin(4x)dx] (which is the antiderivative of sin(4x))..
    Show your work if possible.. Its wrong, but that way we can help correct whatever confusion brought you to that solution.. This is something you'll want to understand fully, as soon as possible.. Calculus is unforgiving to say the least, when your fundamentals aren't REALLY solid..
     
  6. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    This is what you're working with, right?
    \intsin^{4}(x) dx

    mik3 asked you if the integrand was sin(4x), and you mistakenly (I think) said that it was.

    Anyway, the antiderivative is 1/32 * sin(4x) - 1/4 * sin(2x) + 3/8 * x + C, according to a math program I have.

    To get this answer, you need to do a couple of substitutions using these trig identities:
    sin^{2}(x) = (1 - cos(2x))/2 --- (1)
    cos^{2}(x) = (1 + cos(2x))/2 --- (2)

    Using a slightly modified form of the first identity above, you can get:
    sin^{4}(x) = (1 - cos^{2}(2x))/2

    Now, use the second identity to rewrite the right side of the equation just above, and you should be able to get the antiderivative.
    Mark
     
  7. learninmath

    Thread Starter New Member

    May 19, 2008
    5
    0
    ok i see what i did wrong now thankyou,
    this is how i worked it out,
    integral(1+sin4x)/4dx
    =integral 1/2dx+integral cos4xdx (antiderivative of sin being cos)
    =1/2x+1/4cos4x+c
     
  8. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    Nope, that's not right. To convince you, I'll take the derivative of what you ended up with:
    d/dx[x/2 + (cos(4x))/4 + c]
    = 1/2 -sin(4x), which is not the same as (1 + sin(4x))/4

    Also, the derivative of sin(x) is cos(x), but the antiderivative of sin(x) is -cos(x).

    In any case, you're not working the same problem you started this thread with, which I've copied from your original post:
    From what you wrote, I believe the problem should be stated "find the antiderivative of sin^{4}x."
     
  9. learninmath

    Thread Starter New Member

    May 19, 2008
    5
    0
    ok i went to see my tutorer and i got the answer: 1/32 12x-8sin(2x)+sin(4x)

    but im still a bit fuzzy on this like where did the 1/32 come from and what happened to the -cos?

    sorry i still dont get this
     
  10. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    First, the answer you show doesn't even make any sense, because you don't show any operation between 1/32 and 12x.
    Second, this is not the right answer to the problem I believe you are doing, based on your first post.

    Part of the reason you're fuzzy about things is that you don't seem to be clear on exactly what the problem is. This thread has gone off on some unproductive tangents because it's not clear to those reading this thread whether you are trying to find the antiderivative of sin^{4}x or sin (4x), and you haven't cleared that up. In your first post you said sin^4x, which I assume you mean the fourth power of sin(x), which is usually written this way: sin^{4}x.

    Look back at message #6 in this thread, where I gave you the antiderivative and told you how you could get it yourself.
    I have also arrived at this result using the technique I gave in my previous message, so I'm hopeful that if you will reread what I said, your fuzziness will dissipate.
    Mark
     
    Last edited: May 20, 2008
Loading...