Toronto Math Forum
APM3462016F => APM346Lectures => Chapter 2 => Topic started by: Tianyi Zhang on October 06, 2016, 10:40:50 PM

In question 1, I have two integration problems.
(1): With condition line (4), first by using the d'Alembert I got:
$$\frac{1}{2c}\int_{0}^{t} \int_{xc(tt')}^{x+c(tt')}\sin(\alpha x')sin(\beta t')dx'dt'$$
To integrate with respect to $x'$ is easy, I got this:
$$\frac{1}{2\alpha c}\int_{0}^{t}[\cos(\alpha x+\alpha ct\alpha ct')\sin(\beta t')\cos(\alpha x\alpha ct+\alpha ct')\sin(\beta t')]dt'$$
I don't know how to integrate with respect to $t'$ since now I have a mixture of $\sin$ and $\cos$ and they are both about $t$'.
(2): With condition line (7), first by using the d'Alembert I got:
$$\frac{1}{2c}\int_{0}^{t}\int_{xc(tt')}^{x+c(tt')} F'''(x')t' dx'dt'$$
Integrate with respect to x', I got:
$$\frac{1}{2c}\int_{0}^{t}[F''(x+ctct')F''(xct+ct')]t'dt'$$
I don't know how to integrate with respect to $t'$ either.

If you want to integrate a product of $\sin $ or $\cos$ and (another) $\sin $ or $\cos$, use
\begin{align*}
&\cos(\alpha)\cos(\beta)=\frac{1}{2}[\cos (\alpha\beta)+\cos(\alpha+\beta)],\\
&\sin(\alpha)\sin(\beta)=\frac{1}{2}[\cos (\alpha\beta)\cos(\alpha+\beta)],\\
&\sin(\alpha)\cos(\beta)=\frac{1}{2}[\sin (\alpha\beta)+\sin(\alpha+\beta)],
\end{align*}
and BTW, $2+2=4$, if you forgot :D

Thank you, now I get it.