# Homework Help: Integration by parts with orthogonality relation

1. Apr 16, 2014

### FatPhysicsBoy

1. The problem statement, all variables and given/known data

I want to integrate $$\int_{0}^{a} xsin\frac{\pi x}{a}sin\frac{\pi x}{a}dx$$

2. Relevant equations

I have the orthogonality relation:

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

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

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

I know how to do this using a trigonometric identity to swap out the $sin^{2}x$ term for a linear term involving $cos2x$, 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. Apr 16, 2014

### 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.

3. Apr 16, 2014

### pasmith

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}

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

If $dv = \sin(\pi x/a)\sin(\pi x/a)$ then
$$v = \int \sin(\pi x/a)\sin(\pi x/a)\,dx.$$ The integral is indefinite, not a definite integral from $0$ to $a$, so you can't use the orthogonality relation to evaluate it.

4. Apr 16, 2014

### FatPhysicsBoy

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. Apr 16, 2014

### Saitama

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. Apr 16, 2014

### pasmith

We don't need them. If $C$ is a constant then $(v(x) +C)' = v'(x)$ and
$$\int u(x)v'(x)\,dx = u(x)(v(x) + C) - \int u'(x)(v(x) + C)\,dx \\ = u(x)v(x) + Cu(x) - \int u'(x)v(x)\,dx - \int Cu'(x)\,dx \\ = u(x)v(x) - \int u'(x)v(x)\,dx + C\left( u(x) - \int u'(x)\,dx\right) \\ = u(x)v(x) - \int u'(x)v(x)\,dx + C\left( u(x) - (u(x) + D)\right) \\ = u(x)v(x) - \int u'(x)v(x)\,dx - CD$$ and we can combine $-CD$ with the constant we get from integrating $u'v$.

7. Apr 16, 2014

### jaytech

You're answer is wrong because $$\int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \begin{cases} \frac{x}{2} &\mbox{if } n = m; \\ 0 & \mbox{otherwise.} \end{cases}$$

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