Convolution integral $\int_0^t \cos(t-s)\sin(s)\ ds$ How can I calculate the following integral?
$$\int_0^t \cos(t-s)\sin(s)\ ds$$
I can't get the integral by any substitutions, maybe it is easy but I can't get it.
 A: Well, you can use the formula $$\cos\phi\sin\psi=\frac{\sin(\phi+\psi)-\sin(\phi-\psi)}2,$$ which should make things easier, since then $$\cos(t-s)\sin s=\frac12(\sin t-\sin(t-2s)=\frac12\sin t+\frac12\sin(2s-t).$$ At that point, you can split the integral into two parts to evaluate separately.
A: Hint: $\cos(t-s)=\cos t\cos s+\sin t\sin s$, so that the integrand becomes
$$\cos(t-s)\sin(s)=\cos t\cos s\sin s+\sin t\sin s \sin s=\left(\frac{1}{2}\cos t\right)\sin 2s+\sin t \sin^2 s$$
You should be able to integrate this.
A: If you expand the $\cos $ function, you get two integrals, one is of the form
$$
\int \cos s \sin s ds  =-\int \cos s d (\cos s) 
$$
and the other one is 
$$
\int \sin^2 s ds = \int \frac{1- \cos 2 s ds}{2}
$$
which are very simple
A: Let
$$
\mathcal{I}=\int_0^t \cos(t-s)\sin(s)\ ds,\tag1
$$
then using property
$$
\int_b^a f(x)\ dx=\int_b^a f(a+b-x)\ dx,
$$
we have
$$
\mathcal{I}=\int_0^t \cos(s)\sin(t-s)\ ds.\tag2
$$
Adding $(1)$ and $(2)$, we obtain
\begin{align}
2\mathcal{I}&=\int_0^t [\cos(t-s)\sin(s)+\cos(s)\sin(t-s)]\ ds\\
&=\int_0^t \sin(s+t-s)\ ds\\
&=\int_0^t \sin (t)\ ds\\
&=\sin(t)\int_0^t\ ds\\
\mathcal{I}&=\large\color{blue}{\frac{1}{2}t\sin (t)}.
\end{align}
