Author Topic: Question 1 in section 2.4  (Read 216 times)

Tianyi Zhang

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Question 1 in section 2.4
« 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:
$$\frac{1}{2c}\int_{0}^{t} \int_{x-c(t-t')}^{x+c(t-t')}\sin(\alpha x')sin(\beta t')dx'dt'$$
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:
$$\frac{1}{2c}\int_{0}^{t}\int_{x-c(t-t')}^{x+c(t-t')} F'''(x')t' dx'dt'$$
Integrate with respect to x', I got:
$$\frac{1}{2c}\int_{0}^{t}[F''(x+ct-ct')-F''(x-ct+ct')]t'dt'$$
I don't know how to integrate with respect to $t'$ either.
« Last Edit: October 07, 2016, 05:26:26 PM by Tianyi Zhang »

Victor Ivrii

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Re: Question 1 in section 2.4
« Reply #1 on: October 07, 2016, 02:48:02 AM »
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
« Last Edit: October 07, 2016, 08:46:28 PM by Victor Ivrii »

Tianyi Zhang

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Re: Question 1 in section 2.4
« Reply #2 on: October 07, 2016, 05:11:51 PM »
Thank you, now I get it.