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One trigonometric integral

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data
    This problem is solved in my book however I dont totally understand it.

    find the integral of sin^2 (x)

    2. Relevant equations



    3. The attempt at a solution[/

    the integral of sin^2 (x) = integral of (1-cos2x)/2 dx

    = (1/2)x -(1/4)sin2x + C


    I dont understand the las step basically how did the integra of (1-cos2x)/2 was changed into 2 integrals?

    Was it done this way (1-cos2x)/2 = 1/2 (-cos2x) ?


    Thanks a lot in advance.
     
  2. jcsd
  3. Apr 23, 2010 #2

    Char. Limit

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    Actually, your last step is wrong... if you factor out the one half, you get this.

    [tex]\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx = \int \frac{1}{2} - \frac{cos(2x)}{2} dx[/tex]

    Which is solvable. Basically, your quote above...

    Should be...

    See the difference?
     
  4. Apr 23, 2010 #3
    You actually could have just factored out the 1/2 like this:
    [tex]\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx =-\frac{1}{2} \int{cos(2x)} dx[/tex]

    It simplifies the integral.
     
  5. Apr 23, 2010 #4


    so it is the integral of 1/2 - (cos2x)/2

    Now how can I find the integral of - (cos2x)/2 ?

    I have been trying to solve that by parts but I had to solve for more than 4 integrals.
    I choosed u= cos 2x dv = 1/2 dx du -sin2x dx v = 1/2 x
     
  6. Apr 23, 2010 #5
    He made a mistake, he added an additional 1/2 into the integral. The simplified integral is just cos(2x).
     
  7. Apr 23, 2010 #6
    :redface:

    I made a little mistake my self.

    It should be, [
    [tex]\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx =-\frac{1}{2} \int\frac{1}{2}-{cos(2x)} dx [/tex]

    The 1/2 couldn't have been factored out like that, I though it was multiplication. Sorry, if I mess you up.

    Now that we have that established, the integral can be done like this.

    [tex]\int\frac{1}{2} dx-\int{cos(2x)} dx [/tex]
     
  8. Apr 23, 2010 #7
    No problem, im trying to solve for cos2x but I think im doing something wrong the integrals that I solved gave me a messier problem.

    Im doing u = cos2x dv = dx du = -2sen2x v=x

    the result of that was x cos2x + the integrl of 2x sen2x and the result becomes messier when I try to solve for that. what am I doing wrong?
     
  9. Apr 23, 2010 #8

    Dick

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    Stratosphere still doesn't have it right. There is a factor of 1/2 in front of the second integral. And to integrate cos(2x) just do the u-substitution u=2x. No need for parts.
     
  10. Apr 23, 2010 #9
    so it would be (1 - cos2x)/2 = 1/2 - (1/2) cos2x

    Thanks a lot...
     
  11. Apr 23, 2010 #10
    I did it out myself and I got the right answer. I didn't tell him that he had to add a constant of tow into the second integral and then balance it out with a 1/2 in front of it. I though he was suppose to solve it.
     
  12. Apr 23, 2010 #11

    Char. Limit

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    Um, Stratosphere, I really did have the right answer the whole time... The integral you showed had [tex]\int cos(2x) dx[/tex] with no coefficient of 1/2 in front of it. This isn't so. That coefficient needs to be there. If you multiply the inside integral by two, you need to divide the outside by two... To get a coefficient of 1/4 outside of the u-subbed integral.

    I know my multiplication rules, and the distributive property needs to be applied. Both integrals are multiplied by 1/2.
     
  13. Apr 23, 2010 #12
    Well, either way, I got the right answer my self, I don;'t really know how to explain it though.

    And sorry about that, I make little stupid mistakes, (it was right when I woke up).
     
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