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Integration problem

  1. Jan 30, 2006 #1
    Use the identity Sin[x]^2 = [1-Cos[2x]]/2
    to help calculate integral from 0 to t Sin[x]^2 dx
    This question seems really easy but im having some difficulty with it.
    This is what im thinking :
    first put the [1-Cos[2x]]/2 instead of the Sin[x]^2
    so this is what i have now (int from 0 to t) [1-Cos[2x]]/2dx
    then i thought let u = cos[2x] then du = - Sin[2x]dx
    and when x= 0 then u = 1 and when x= t then u= cos[2t]
    then (int from 0 to t) [1-Cos[2x]]/2dx
    =-1/2 (int from 1 to cos[2t] ) [1-u]/[Sin[2x]du]
    Now i dont know what to do please help!!!:confused: :surprised
  2. jcsd
  3. Jan 30, 2006 #2
    Why did you make a substitution after you used the trig identity to get rid of the sine squared, cos(2x) is certainly integrable.
  4. Jan 30, 2006 #3
    back to the integration problem

    the substitution was the first thing i did i should leave the sin[x]^2 like it is then??
    let sin[x]= u ???
    what is the integral of Cos[2x]??
  5. Jan 30, 2006 #4

    Ok i think i got it:shy:
    I just one clerification is sin^2[x] the same as Sin[x]^2???
  6. Jan 30, 2006 #5


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  7. Jan 30, 2006 #6
    im very sorry ,I'm new ,i didn't know..
  8. Jan 30, 2006 #7
    Yes that's correct. That notation is used to not confuse the following expressions.

    [tex]\sin(x)^2[/tex] and [tex]\sin(x^2)[/tex]
  9. Jan 31, 2006 #8
    not much trouble by the way
    but a slight mistake
    if u=cos2x
    than d(u)=-sin2xd(2x)
    but i dont really know why u r making this substitution?
    cos2x is a basic integration, but if u want to do it this way which is useless than do tell i will tell u than how to do this useless thing
  10. Jan 31, 2006 #9
    u rnt using [x] to denote gretest integer function? that would make it slightly complex
  11. Jan 31, 2006 #10
    No, he's not using [x] as the greatest integer function. Anyway, here's the original integral.

    [tex]\int \sin^2(x)dx=\frac{1}{2} \int 1-\cos(2x)dx= \frac{x}{2}-\frac{\sin(2x)}{4}[/tex]

    No constant of integration because it's a definite integral. Just plug in your bounds.
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