how to calculate the following expression? Let $f$ be a smooth function and $\alpha &gt;0$, i would like to know how to calculate the following expression: $$ A:=\dfrac{d}{dt}\left[\int_0^t f(s) \sin(\alpha(t-s))ds\right] $$
I followed a very simple method but i think that it is wrong.
I assumed that $$ \int_0^t f(s) \sin(\alpha(t-s))ds=G(t)-G(0) $$
where $$G(s)=\int g(s) ds$$ and $$g(s)=f(s)\sin(\alpha(t-s))$$
Then we have $$ A= \frac{d}{dt}\left(G(t)-G(0)\right)= g(t)=0$$
Question:

Did i make any mistakes ?

How to proceed to calculate this type of integral ?
Note that $t$ is also present in the integrand, so you need to apply the chain rule:
$$
\frac{d}{dt}\int_0^t g(t,s)ds=g(t,t)+\int_0^t ds \frac{d}{dt}g(t,s)\ .
$$
In your case, $g(t,t)$ is indeed equal to zero, but the second chunk is not...
$$ A:=\frac{d}{dt}\left[\int_0^t f(s) \sin(\alpha(t-s))ds\right] =$$
$$=\frac{d}{dt}\left[\sin(\alpha{t})\int_0^t f(s)\cos(\alpha{s}) ds-\cos(\alpha{t})\int_0^t f(s)\sin(\alpha{s}) ds\right]=$$
$$=\alpha\cos{\alpha{t}}\int_0^t f(s)\cos(\alpha{s}) ds+\alpha\sin{\alpha{t}}\int_0^t f(s)\sin(\alpha{s}) ds=$$
$$\alpha\int_{0}^{t}f(s)\cos{(\alpha(t-s))}ds$$