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Homework Help: Integral of sin(x) sin(x+1) dx from 0 to 2pi.

  1. Oct 6, 2013 #1


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    1. The problem statement, all variables and given/known data
    Integral of sin(x) sin(x+1) dx from 0 to 2pi.

    2. Relevant equations
    Integration by parts: Integral u dv = uv – Integral v du

    3. The attempt at a solution
    My work has been attached as MyWork.jpg. I, basically, get 0 * integral_I_started_with = something_else instead of nonzero_constant * integral_I_started_with = something_else and, given that I've done this problem a multitude of times and failed (if I remember correctly, always with the same issue) and, I was hoping someone here could point out my mistake.

    Any input would be greatly appreciated!

    Attached Files:

  2. jcsd
  3. Oct 6, 2013 #2


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    Try a different approach. There's a trig formula that will let you express sin(A)sin(B) as a sum of simple trig functions. Can you find it?
  4. Oct 6, 2013 #3


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    That image is difficult to read and in any case you didn't put in the limits. But integration by parts isn't what you want to do anyway. Use the formula$$
    \sin a \sin b =\frac 1 2 (\cos(a-b)-\cos(a+b))$$
  5. Oct 6, 2013 #4

    D H

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    Your double integration by parts yielded
    [tex]\int \sin(x)\sin(x+1)dx = -\sin(x+1)\cos(x) + \cos(x+1)\sin(x) + \int \sin(x)\sin(x+1)dx[/tex]
    (This is the line just before the line that is colored red.) Subtract the integral from both sides and you get
    [tex]0 = -\sin(x+1)\cos(x) + \cos(x+1)\sin(x)[/tex]
    One may rightfully ask, what gives?

    Look at the right hand side, ##-\sin(x+1)\cos(x) + \cos(x+1)\sin(x)##. This is ##\sin(x-(x+1))##, or ##\sin(1)##. So now we have the even more nonsensical ##\sin(1)=0##, which obviously is not true.

    What happened is that you dropped a constant of integration. That supposedly nonsensical ##\sin(1)=0## is better stated as ##\sin(1) = c##. Instead of reaching a contradiction you've reached a tautology.

    Bottom line: That's Just one of those things you need to be aware of when you do integration by parts.

    Here's a hint that will help you solve this integral: Use the trick that x-x=0.
    With this, cos(a) = cos(a+x-x) = cos((a+x)-x) = cos(a+x)cos(x) + sin(a+x)sin(x).
    Now substitute a=1.
  6. Oct 11, 2013 #5


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    Thanks, guys!
  7. Oct 11, 2013 #6
    An alternative approach.
    $$I=\int_{0}^{2\pi} \sin(x)\cdot\sin(x+1)dx $$

    The integral is equivalent to
    $$I=\int_{0}^{2\pi} \sin(2\pi-x)\cdot \sin(2\pi-(x-1))dx=\int_0^{2\pi} \sin(x)\cdot\sin(x-1)dx$$

    Add the two expressions to get,
    $$2I=\int_0^{2\pi} \sin(x)(\sin(x+1)+\sin(x-1))dx$$
    Use ##\sin(A)+\sin(B)=2\sin((A+B)/2)\cos((A-B)/2)## to obtain,
    $$2I=\int_0^{2\pi} \sin(x)(2\sin(x)\cos(1))dx \Rightarrow I=\cos(1)\int_0^{2\pi} \sin^2x dx$$

    From the formula, ##\cos(2x)=1-2\sin^2x##, the integral can be easily evaluated.

    I hope that helped.
  8. Oct 13, 2013 #7
    This problem could have been solved with much more ease, if you multiply the integral and divide by 2 and then isolate 1/2 out. Then apply formula 2sinAsinB=cos(A-B)-cos(A+B).

    I know its too late to post here though.

    Edit: Excellent hint given by LC Kurtz and Dick.
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