Integration by parts with orthogonality relation

  • #1

Homework Statement



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


Homework 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]

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:

Answers and Replies

  • #2
35,287
7,140

Homework Statement



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

Homework 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]

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.
 
  • #3
pasmith
Homework Helper
2,074
695

Homework Statement



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


Homework Equations



I have the orthogonality relation:

[tex]\int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \left \{ \begin{matrix}
\frac{a}{2} \text{, if \textit{n = m};}\\
0 \text{, otherwise.}
\end{matrix}\right.[/tex]

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

and the parts formula:
[tex]\int u \, dv=uv-\int v \, du[/tex]

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.

If [itex]dv = \sin(\pi x/a)\sin(\pi x/a)[/itex] then
[tex]
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.
 
  • #4
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:
I(n,m)  = \begin{cases} \frac{a}{2}, & \mbox{if $n = m$}, \\[1em]
0 & \mbox{else} \end{cases}
gives
[tex]
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
[tex]
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.

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?
 
  • #5
3,816
92
There isn't a need to use integration by parts. You can easily solve this by using the properties of definite integral.

Let:
$$I=\int_0^a x\sin^2\left( \frac{\pi x}{a}\right)\,dx$$
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.
 
  • #6
pasmith
Homework Helper
2,074
695
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?

We don't need them. If [itex]C[/itex] is a constant then [itex](v(x) +C)' = v'(x)[/itex] and
[tex]
\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].
 
  • #7
64
6

Homework 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]

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.
 

Related Threads on Integration by parts with orthogonality relation

  • Last Post
Replies
1
Views
890
  • Last Post
Replies
3
Views
940
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
2
Views
748
Replies
2
Views
696
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
952
Top