Please check my integration

[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.

[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?

[e(ho0n3]

(\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}


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.

[danoonez]

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.


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?

[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.

[danoonez]

Excellent. Thanks for looking.