It is well known that the Dirac delta function has the following property:
$\int_{-\infty}^{\infty}\delta(t-a)f(t)dt=f(a)$
If $g(t)=\int_{0}^{t}\sin(t-\tau)\delta(\tau-\pi)d\tau$
then $g(t) = \left\{ \begin{array}{ll} 0, & \textrm{if}\quad t<\pi\\ \sin(t-\pi), & \textrm{if}\quad t\geq\pi \end{array} \right.$
How can you show this result?
Surely this is no the "optimal" way to show it and a simple shift of variables will do, but...
Since $\delta (\tau -\pi )=\frac{\mathrm d \theta (\tau -\pi )}{\mathrm d \tau }$ (where $\theta (\tau -\pi )$ is the Heaviside theta function) you can set it like that: $$g(t)=\int_{0}^{t}\sin(t-\tau)\delta(\tau-\pi)\mathrm d\tau=\int_{0}^{t}\sin(t-\tau)\mathrm d\theta (\tau -\pi )$$ Then integrate it by parts and use the definition of Heaviside theta function: $$ \require{cancel} \begin{eqnarray} g(t)&=&\cancelto{0}{\sin(t-\tau)\theta (\tau -\pi )\bigg|_0^t}-\int_{0}^{t}\frac{\partial \theta (\tau -\pi )}{\partial \tau }\mathrm d\sin(t-\tau)=\\ &=&-\int_{0}^{t}\theta (\tau -\pi )\mathrm d\sin(t-\tau)=\int_{0}^{t}\cos(t-\tau)\theta (\tau -\pi )\mathrm d\tau=\\ &=&\cases{\int_{0}^{t}\cos(t-\tau)\cdot 0 \ \mathrm d\tau \quad \mbox{if} \quad t<\pi\\\int_{t}^{t}\cos(t-\tau)\cdot \frac{1}{2} \ \mathrm d\tau \quad \mbox{if} \quad t=\pi\\\int_{\pi}^{t}\cos(t-\tau)\cdot 1 \ \mathrm d\tau \quad \mbox{if} \quad t>\pi }\\&=&\cases{0\quad \quad \quad \quad \ \ \mbox{if} \quad t<\pi\\0 \quad \quad \quad \quad \ \ \mbox{if} \quad t=\pi\\\sin(t-\pi) \quad \mbox{if} \quad t>\pi } \end{eqnarray} $$