This is a homework problem, so I'm not looking for a worked out solution, merely to be pointed in the right direction.
Convolve x(t) with h(t) where:
$$
x(t) = sin(t) \\
h(t) = e^{-.1t}sin(2t)u(t)
$$
u(t) is the unit step function. Here is where I got:
$$
y(t) = \int^{\infty}_{-\infty} sin(t – \tau)e^{-.1\tau}sin(2\tau)u(\tau) \, d\tau
$$
$$
y(t) = \int^{\infty}_{0} sin(t – \tau)e^{-.1\tau}sin(2\tau)\, d\tau
$$
At this point I'm pretty lost, I don't recognize any normal techniques of integration that can handle this. After applying some trig identities I get:
$$
y(t) = \frac{1}{2}\int^{\infty}_{0} e^{-.1\tau}cos(t – \tau) \, d\tau – \frac{1}{2}\int^{\infty}_{0} e^{-.1\tau}cos(t) \, d\tau
$$
This seems a little better(I can evaluate the second integral), but I'm still lost on the first. I'd appreciate any help at all.
Best Answer
Hint: $$ \begin{align} \sin(t-\tau)\sin(2\tau) &=\frac{\cos(t-3\tau)-\cos(t+\tau)}{2}\\ &=\frac{\cos(t)\cos(3\tau)+\sin(t)\sin(3\tau)}{2}\\ &-\frac{\cos(t)\cos(\tau)-\sin(t)\sin(\tau)}{2}\tag{1} \end{align} $$
Integration by parts gives $$ \begin{align} \int_0^\infty\cos(ax)\,e^{-bx}\,\mathrm{d}x &=-\frac1b\int_0^\infty\cos(ax)\,\mathrm{d}e^{-bx}\\ &=\frac1b-\frac ab\int_0^\infty\sin(ax)\,e^{-bx}\,\mathrm{d}x\tag{2} \end{align} $$ and $$ \begin{align} \int_0^\infty\sin(ax)\,e^{-bx}\,\mathrm{d}x &=-\frac1b\int_0^\infty\sin(ax)\,\mathrm{d}e^{-bx}\\ &=\frac ab\int_0^\infty\cos(ax)\,e^{-bx}\,\mathrm{d}x\tag{3} \end{align} $$ Combining $(2)$ and $(3)$ yields $$ \int_0^\infty\cos(ax)\,e^{-bx}\,\mathrm{d}x=\frac{b}{a^2+b^2}\tag{4} $$ and $$ \int_0^\infty\sin(ax)\,e^{-bx}\,\mathrm{d}x=\frac{a}{a^2+b^2}\tag{5} $$