The input signal for a given electronic circuit is a function of time $V_{in}(t)$. The output signal is given by
$$V_{out}(t) = \int_0^t \sin(t-s)V_{in}(s)ds$$
Find $V_{out}(t)$ if $V_{in}(t) = \sin(at)$ where $a > 0$ is some constant
I believe that this can simplify to the integral of $\displaystyle\frac{1}{2}(\cos(t-(a+1)s)-\cos(t+(a-1)s))$ taken from $0$ to $t$, but I do not know how to continue from here.
$$V_{out}(t)=\int_0^1\sin(t-s)V_{in}(s)ds$$ $$V_{in}(t)=\sin at$$ $$\begin{align}V_{out}(t)&=\int_0^1(\sin(t-s)\sin (as))ds\\ &=\int_0^1(\sin t\cos s\sin as-\cos t\sin s\sin as)ds\\ &=\sin t\int_0^1(\sin as\cos s)ds-\cos t\int_0^1(\sin s\sin as )ds\end{align}$$ Now use $$2\sin x\sin y=\cos(x-y)-\cos(x+y),2\sin x\cos y=\sin(x+y)+\sin(x-y)$$ So $$\begin{align}2V_{out}(t)&=\sin t\int_0^1(\sin (a+1)s+\sin(a-1)s)ds-\cos t\int_0^1(\cos(a-1)s-\cos(a+1)s)ds\\ &=\sin t\left[-\frac{\cos(a+1)s}{a+1}-\frac{\cos(a-1)s}{a-1}\right]_0^1-\cos t\left[\frac{\sin(a-1)s}{a-1}+\frac{\sin(a+1)s}{a+1}\right]_0^1\\ &=\cdots\end{align}$$ So: $$V_{out}(t)=\frac{a(\cos a\sin(1-t)+\sin t)-\sin a\cos(1-t)}{a^2-1}$$
