Can anyone help with the following integral:
int[sqrt(2/L)*sin((n*Pi*x)/L)]dL
The hint given is to use the trig identity:
sin(alpha)*sin(beta) = (1/2)*cos(alpha - beta) - (1/2)*cos(alpha + beta)
Thanks.
Hello, Ideasman!
If no one is responding, it's that the problem makes no sense.
. . I did an autopsy and came up with a COD . . .
int[sqrt(2/L)*sin((n*Pi*x)/L)]dL . ??
The hint given is to use the trig identity:
. . $\displaystyle \sin(\alpha)\cdot\sin(\beta) = \frac{1}{2}\cos(\alpha - \beta) - \frac{1}{2}\cos(\alpha + \beta)$
Could this be the problem?
. . $\displaystyle \int\sin\left(\frac{2}{L}\right)\cdot\sin\left(\fr ac{n\pi}{L}x\right)\,dx$
. . . . .$\displaystyle \uparrow$ . . . . . . . . . . . . $\displaystyle \uparrow$
Apply the identity:
. . $\displaystyle \sin\left(\frac{2}{L}\right)\cdot\sin\left(\frac{n \pi}{L}x\right) \:=\:\frac{1}{2}\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right) - \frac{1}{2}\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)$
The problem becomes:
. . $\displaystyle \frac{1}{2}\int\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right)\,dx - \frac{1}{2}\int\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)\,dx$
Then use substitution: .$\displaystyle u \:=\:\frac{2}{L} - \frac{n\pi}{L}x$ and $\displaystyle v \:=\:\frac{2}{L} + \frac{n\pi}{L}x$
It's a good thought. I came up with something similar, but I think we would have to leave the integral as "dL" otherwise the first sin function is simply a constant and we wouldn't need the hint. In fact, the integral becomes rather trivial. The problem I had with this is if we are integrating over L then I can't find a way to do the integral, and I suspect it has no closed form. (If I remember correctly anyway.)
But you are definitely right about one thing: this problem makes no sense as it is stated.
-Dan
Hello, Dan!
I didn't notice that the first sine is constant . . . *blush*
You're right . . . with $\displaystyle dL$, we can't integrate it:
. . $\displaystyle \frac{1}{2}\int\cos\left[(2-n\pi x)L^{-1}\right]\,dL - \frac{1}{2}\int\cos\left[(2+n\pi x)L^{-1}\right]\,dL $
That's what convinced me that it must be $\displaystyle dx$.
I hope Ideasman clarifies it soon . . .