I am trying to understand a separate way of showing $\mathcal{L}\{sin(at)\}$ using the the following method:
What I have so far is
$$\text{Let $g(t)=f(at)$ where we substitute $u=at$. Then we have} $$ $$\mathcal{L}\{g(s)\}=\int_0^\infty e^{-st} g(t)\;dt$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\int_0^\infty e^{-st} f(at) \;dt$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{a}\ \int_0^\infty e^{-st} f(u) \;du$$ $$\;\;\;\;\;\;= \frac{1}{a} \mathcal{L}\{f(u)\}$$ $$\;\;\;\;\;\;\;= \frac{1}{a} \mathcal{L}\{f(at)\}$$
From this formula and the fact that $\mathcal{L}\{sin(t)\}=\frac{1}{1+s^2}\;$, I then solve this to be:
$$\mathcal{L}\{sin(t)\}= \frac{1}{a}\:\mathcal{L}\{sin(at)\}$$ $$therefore \;\;\mathcal{L}\{sin(at)\}=\frac{a}{1+s^2}$$
Clearly, there is an issue since $sin(at) = \frac{a}{a^2+s^2}$, so I was wondering where I might be making a mistake? Thank you!
You were almost there just a little mistake..
You need to substitute also t in the exponential $$\mathcal{L}(f(at))=\frac{1}{a}\ \int_0^\infty e^{-\color{red}{st}} f(u) \;du$$ you should have $$\mathcal{L}(f(at))=\frac{1}{a}\ \int_0^\infty e^{-su/a} f(u) \;du=\frac 1 aF\left(\frac s a\right)$$ $$\implies \mathcal{L}(\sin(at))=\frac 1a \frac 1 {1+s^2/a^2}$$ $$\mathcal{L}(\sin(at))=\frac a {a^2+s^2}$$