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

This looks like an easy integral, why cant i get it

  1. Jul 6, 2005 #1
    the integral of sin(x)^2

    I tried using all kinds of trigonometric identities, but they never seemed to make it easier. Any suggestions on what identities to use, or how to approach this? Thanks
  2. jcsd
  3. Jul 6, 2005 #2


    User Avatar
    Homework Helper

  4. Jul 7, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    If it's [itex] \sin x^{2} [/itex] the function you wish to antidifferentiate, you can express the result in terms of a Fresnel integral, "S" if i'm not mistaking.

  5. Jul 8, 2005 #4
    int of sin ^2 (x) dx = 1/2 x - 1/4 sin 2x + c
  6. Jul 8, 2005 #5
    You're not supposed to give away the answer... tsk tsk
  7. Jul 13, 2005 #6
    ru talking of sin(x)sin(x) or is it sin (x*x)??? the 1st one seems quite easy...1-cos(2x) / 2 wud do it....for the second one...integrate by parts taking x^0 as the second function...then u will get sumthing like integral of x^2cos(x^2) in the second integral...integrate partially again...u will get LHS with a minus sign on the RHS(check it out...im not very sure...i did it in rough ;)) ... bring it over to LHS and finish it...check it out...im not so sure of my arithmetic.
  8. Jul 13, 2005 #7
    I'm fairly sure that the second interpretation cannot be integrated by parts. As Daniel said, it requires the Fresnel integral, which integrals.com confirmed.

    [tex]S_1(x) = \sqrt{\frac{{2}}{{\pi}}}\int_{0}^{t}\sin{x^2}dx[/tex]

    I believe this is the one.
    Last edited: Jul 13, 2005
  9. Jul 13, 2005 #8

    Is this methodology correct?

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

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

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

    [tex]\frac{1}{2} \left( \int dx - \int \cos (2x) dx \right) = \frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C[/tex]

    [tex]\frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C = \frac{x}{2} - \frac{\sin (2x)}{4} + C[/tex]

    [tex]\boxed{\int \sin^2 x \; dx = \frac{x}{2} - \frac{\sin (2x)}{4} + C}[/tex]
    Last edited: Jul 14, 2005
  10. Jul 13, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    You should always insert the additive constant in an indefinite integral.

  11. Jul 13, 2005 #10
    Yes, the "+ C" is needed, but besides that you are correct. And you had a correct approach. That's a common Calc I question I believe.
  12. Jul 13, 2005 #11


    User Avatar
    Science Advisor
    Homework Helper

    try parts. thats my favorite way.
  13. Jul 14, 2005 #12

    Mathematica Fresnel Integral definition:
    [tex]S(z) = \int_0^z \sin \left( \frac{\pi t^2}{2} \right) dt[/tex]

    [tex]\int \sin (x^2) dx = \sqrt{\frac{2}{\pi}} \int_0^t \sin (x^2) dx[/tex]

    integrals.wolfram.com solution:
    [tex]\boxed{\int \sin (x^2) \; dx = \sqrt{\frac{\pi}{2}} \cdot \text{FresnelS} \left[ \sqrt{\frac{2}{\pi}} x \right]}[/tex]

    Is this methodology correct? and what intermediate steps are missing?

    What is the integrals.wolfram.com mathematical solution written in symbolic form?

    Last edited: Jul 14, 2005
  14. Jul 14, 2005 #13


    User Avatar
    Science Advisor
    Homework Helper

    for the integral sin^2(x) parts gives

    int(sin^2(x)dx) = -cos(x)sin(x) + int(cos^2(x)dx)

    = -cos(x)sin(x) + int(1 - sin^2(x)dx)

    so 2 int(sin^2(x)dx) = x - cos(x) sin(x) +constant.

    this way you do not need to have any tricky formulas at hand like

    sin2 = (1/2)(1-cos(2x)), and the answer comes slightly simpler too.

    sin(x^2) on the other hand is no easier than the famous e^(x^2).

    intuition should suggest (by the product rule) that most functions of form

    f(g(x)) that cannot be expanded or simplified are not going to occur as elementary derivatives.

    with a bit of work one can do sqrt(1+x^2) but only because the substitution x = tan(u) simplifies it.

    one cannot do sqrt(1+x^4) this way though. :rofl:
    Last edited: Jul 14, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook