1. Oct 4, 2004

danoonez

From

$$\int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1$$

show that

$$A = \sqrt {\frac {2} {L}$$

Here's what I did:

First bring $$A^2$$ out so that I have:

$$A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1$$

Then I use u subsitution for the integral:

$$\int sin^2 \frac {n \pi x} {L} dx$$

equals:

$$\int \frac {L} {n \pi} sin^2 u du$$

Pulling out the constants I get:

$$\frac {L} {n \pi} \int sin^2 u du$$

Which, using an integral table, equals:

$$(\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)$$

Plugging u back in and putting in my limits of integration I get my definite integral:

$$(\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L}$$

Plugging this back into the original equation I get:

$$A^2 (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L} = 1$$

Then solving for the limits of integration I get:

$$A^2 (\frac {L} {n \pi}) \left[ (\frac {n \pi L} {2L} - \frac {sin 2 n \pi L} {4 L}) - (0 - 0) \right] = 1$$

Then doing some algebra I get:

$$A^2 (\frac {n \pi L^2} {2 L n \pi} - \frac {sin 2 n \pi L^2} {4 L n \pi}) = 1$$

...then some scratch work...

$$sin \pi = 0$$

...so...

$$(A^2) (\frac {n \pi L^2} {2 L n \pi}) = 1$$

After more algebra I get:

$$(A^2) (\frac {L} {2}) = 1$$

So then:

$$A^2 = \frac {2} {L}$$

And finally:

$$A = \sqrt {\frac {2} {L}$$

Q.E.D.

How does it look.
Sorry if it seems messy; this was my first time using LaTex.

2. Oct 4, 2004

vsage

There's a step around plugging n*pi*x/L back in for u where inexplicably the L in n*pi*x/L was pulled out of the sine function. Was this a typo?

3. Oct 4, 2004

e(ho0n3

You should really put parentheses around the variables in the sine function, i.e.
$$\Big(\frac {L} {n \pi}\Big) \left[ \frac {n \pi x} {2L} - \frac {\sin (2 n \pi x/L)} {4} \right]_{0}^{L}$$

Everything else looks OK.

4. Oct 4, 2004

danoonez

Thank you. I will do that for the paper I hand in.

vsage: I can't tell which step you mean. Could you clarify? B/c it may not have been a typo.

Anybody else see any mistakes?

5. Oct 4, 2004

vsage

I was referring to what e(ho0n3 pointed out. You seem to account for the typo though later so it looks great to me.

6. Oct 4, 2004

danoonez

Excellent. Thanks for looking.