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Homework Help: Integral (sin x/2 - cos x/2)^2

  1. Dec 26, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm unable to solve this integral. I get a result, but it doesn't match the solution.

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

    3. The attempt at a solution
    CodeCogsEqn_gif_300x300_q85.jpg
    (attachment?)
    I'd be grateful for highligting my errors.
     

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  2. jcsd
  3. Dec 26, 2011 #2
    You didnt do the substitution properly.
    And there is no need for substitution. Note the identity: [itex]\sin(2x)=2\sin x \cos x[/itex]
     
  4. Dec 26, 2011 #3
    2sin(x)cos(x) = sin(2x)dx

    ∫ [2sin(x)] [cos(x)dx] = sin2(x)
    ∫ [-2cos(x)] [-sin(x)dx] = -cos2(x)
    ∫ sin(2x)dx = -(1/2)cos(2x)
     
  5. Dec 26, 2011 #4

    Curious3141

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    Hint: simplify with the double-angle formula first.
    sin(2z) = 2(sinz)(cosz)

    Now put z = (1/2)x and see what you get on the RHS. What do you get on the LHS? Integrate that.
     
  6. Dec 26, 2011 #5

    SammyS

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    Hello b0rsuk. Welcome to PF !

    What is the correct answer?
     
  7. Dec 26, 2011 #6
    Hmm. In such case,
    [tex]2 \int \sin \frac{x}{2} \cos \frac{x}{2} dx = \int 2 \sin \frac{x}{2} \cos \frac{x}{2} dx = \int \sin x dx = -\cos x + C_1[/tex]

    But what about the first integral ? I know:
    [tex]\sin^2 x + \cos^2 x = 1[/tex]
    But I have:
    [tex]\int\sin^2 \frac{x}{2} + \cos^2\frac{x}{2} dx[/tex]
    Can I simply get around that with substitution ? Say,
    [tex]\frac{x}{2} = t, \frac{1}{2} = dt[/tex]
    Then I get
    [tex]\frac{1}{2}(\sin^2 t + \cos^2 t)[/tex]
    Does it equal 1 (or, actually, 1/2) ?
    --------------------------

    The solution, according to the book:
    [tex]x + \cos x + C[/tex]
    Yes, that's a plus.

    ---------------
    You people are scary. I'm a CS graduate and I came here to defeat my arch-nemesis, math. This time, without time pressure. You rob me of any excuses I might have to procrastinate solving examples.
     
  8. Dec 26, 2011 #7

    Curious3141

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    [itex]\sin^2x + \cos^2x = 1[/itex] is an identity. Meaning the value of x can be *anything*. Even if you replace the x with x/2, it still holds (just to be clear, that means [itex]\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} = 1[/itex]). Don't bother with substitution, etc. Just recognise this and reduce the integrand to 1 first.

    So the first integrand is 1, and the integral is x.

    The second integrand is [itex]-\sin x[/itex] (you correctly reflected the minus sign in your first post, but then you must have forgotten it in your most recent one). The integral is [itex]\cos x[/itex].

    So the final answer is [itex]x + \cos x + C[/itex].
     
    Last edited: Dec 26, 2011
  9. Dec 26, 2011 #8

    D H

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    In this case, simple trig substitutions are all you need.
    [tex]\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1 - \sin x[/tex]
    This is an identity, so [itex]\int (\sin x/2 - \cos x/2)^2 dx = \int (1-\sin x)\,dx[/itex]. The right hand side obviously integrates to [itex]x+\cos x + c[/itex].

    So how to arrive at that identity?


    One way is to expand the square,
    [tex]\left(\sin\frac x 2 - \cos \frac x 2\right)^2 =
    \sin^2 \frac x 2 - 2 \sin \frac x 2 \cos \frac x 2 + \cos^2 \frac x 2[/tex]
    Now use the identities [itex]\sin^2 \theta + \cos^2\theta = 1[/itex] and [itex]\sin(2\theta) = 2 \sin\theta\cos\theta[/itex] to arrive at the desired identity
    [tex]\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1 - \sin x[/tex]

    Another way is to use the identity [itex]\sin\theta- \cos \theta = \sqrt 2 \sin(\theta-\pi/4)[/itex]. This comes in handy at times, but doesn't quite fall into the camp of identities you should just "know". With this,
    [tex]\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 2\sin^2\left(\frac x 2 - \frac \pi 4\right) = 2\sin^2\left(\frac 1 2(x-\pi/2)\right)[/tex]
    Now use the half-angle formulae [itex]2\sin^2(u/2) = 1-\cos u[/itex]. The half-angle formulae come into play quite often. With this,
    [tex]\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1-\cos\left(x-\frac{\pi} 2\right) = 1 - \sin x[/tex]
     
  10. Dec 26, 2011 #9

    SammyS

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    After splitting this up into two integrals, the first one is fine.

    You then made offsetting errors -- or simply had a typo.
    [itex]\displaystyle -2\int\left(\sin\left(\frac{x}{2}\right)\cos\left( \frac{x}{2}\right)\right)dx=-4\int\left(\frac{1}{2}\sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)\right)dx[/itex]
    [itex]\displaystyle=-4\int\left(\sin(t)\cos(t)\right)dt[/itex]​
    After that you dropped a minus sign.

    So, the final answer should be [itex]\displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C[/itex]

    I know that's not the book's answer, but notice that
    [itex]1-2\sin^2(\theta)=\cos(2\theta)[/itex]

    So substituting x/2 for θ and doing a bit of algebra gives

    [itex]\displaystyle -2\sin^2\left(\frac{x}{2}\right)=\cos(x)-1[/itex]​

    Therefore, [itex]\displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C=x+\cos(x)-1+C[/itex]

    -1+C is just a constant, one unit less than C.

    So, your answer was correct except for a sign error.
     

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