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Homework Help: Integration by parts with orthogonality relation

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data

    I want to integrate [tex]\int_{0}^{a} xsin\frac{\pi x}{a}sin\frac{\pi x}{a}dx[/tex]

    2. Relevant equations

    I have the orthogonality relation:

    [tex]\int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \begin{cases} \frac{a}{2} &\mbox{if } n = m; \\
    0 & \mbox{otherwise.} \end{cases} [/tex]

    and the parts formula:
    [tex]\int u \, dv=uv-\int v \, du[/tex]
    3. The attempt at a solution

    I took [itex]u = x[/itex], and [itex] dv = sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}[/itex]. Following the parts formula I get a final answer of [itex]\frac{a^{2}}{2} - \frac{a^{2}}{2} = 0[/itex]. However, this is incorrect. The correct answer for the integral is [itex]\frac{a^{2}}{4}[/itex].

    I know how to do this using a trigonometric identity to swap out the [itex]sin^{2}x[/itex] term for a linear term involving [itex]cos2x[/itex], but I don't quite understand why this method doesn't work. I think it has something to do with linearity but I don't fully understand why one method works and the other doesn't.
    Last edited: Apr 16, 2014
  2. jcsd
  3. Apr 16, 2014 #2


    Staff: Mentor

    This is the same as
    $$ \int_{0}^{a} xsin^2(\frac{\pi x}{a})dx$$

    An easier way to integrate this is to use a trig double angle formula.
  4. Apr 16, 2014 #3


    User Avatar
    Homework Helper

    If you have a "\left", you need a matching "\right". "\right." is necessary if you don't want a closing symbol. But this sort of thing is best done using "\begin{cases}":
    Code (Text):

    I(n,m)  = \begin{cases} \frac{a}{2}, & \mbox{if $n = m$}, \\[1em]
    0 & \mbox{else} \end{cases}
    I(n,m) = \begin{cases} \frac{a}{2}, & \mbox{if $n = m$} \\[1em]
    0 & \mbox{else} \end{cases} [/tex]

    If [itex]dv = \sin(\pi x/a)\sin(\pi x/a)[/itex] then
    v = \int \sin(\pi x/a)\sin(\pi x/a)\,dx.
    [/tex] The integral is indefinite, not a definite integral from [itex]0[/itex] to [itex]a[/itex], so you can't use the orthogonality relation to evaluate it.
  5. Apr 16, 2014 #4
    Thanks for that, I managed to get it sorted in the end, I hit 'Submit Reply' by accident instead of 'Preview Post'.

    I suppose that does make sense when I consider all the other instances where I've used product rule. We never evaluate the v integrand definitely. If the integral is always indefinite then why don't we have constants of integration floating about when we do integration by parts?
  6. Apr 16, 2014 #5
    There isn't a need to use integration by parts. You can easily solve this by using the properties of definite integral.

    $$I=\int_0^a x\sin^2\left( \frac{\pi x}{a}\right)\,dx$$
    The above is same as:
    $$I=\int_0^a (a-x)\sin^2\left( \frac{\pi x}{a}\right)\,dx$$
    Add the two expressions for ##I## to get:
    $$2I=\int_0^a a\sin^2\left( \frac{\pi x}{a}\right)\,dx$$
    Use the orthogonality relation you posted to get the required answer.
  7. Apr 16, 2014 #6


    User Avatar
    Homework Helper

    We don't need them. If [itex]C[/itex] is a constant then [itex](v(x) +C)' = v'(x)[/itex] and
    \int u(x)v'(x)\,dx = u(x)(v(x) + C) - \int u'(x)(v(x) + C)\,dx \\
    = u(x)v(x) + Cu(x) - \int u'(x)v(x)\,dx - \int Cu'(x)\,dx \\
    = u(x)v(x) - \int u'(x)v(x)\,dx + C\left( u(x) - \int u'(x)\,dx\right) \\
    = u(x)v(x) - \int u'(x)v(x)\,dx + C\left( u(x) - (u(x) + D)\right) \\
    = u(x)v(x) - \int u'(x)v(x)\,dx - CD
    [/tex] and we can combine [itex]-CD[/itex] with the constant we get from integrating [itex]u'v[/itex].
  8. Apr 16, 2014 #7
    You're answer is wrong because [tex]\int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \begin{cases} \frac{x}{2} &\mbox{if } n = m; \\
    0 & \mbox{otherwise.} \end{cases} [/tex]

    The [itex]\frac{a}{2}[/itex] is after evaluation of the integral from 0 to a. So your [itex]\int[/itex]vdu is actually [itex]\int[/itex][itex]\frac{x}{2}[/itex]dx.
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