Angular impulse:

If \( \vec{\tau} \) be the total external torque applied to a system of particles about the origin, then the total angular impulse of the system will be \( \displaystyle{\int_{t_1}^{t_2}}\vec{\tau}dt \).

We know that \( \vec{\tau}=\frac{d\vec{L}}{dt} \), [ to know the derivation (CLICK HERE) ]

If \( \vec{L_1} \) and \( \vec{L_2} \) be the total angular momenta of the system at times \( t_1 \) and \( t_2 \) respectively, then

\( \displaystyle{\int_{t_1}^{t_2}}\vec{\tau}dt=\displaystyle{\int_{t_1}^{t_2}}\frac{d\vec{L}}{dt}dt\\=\displaystyle{\int_{t_1}^{t_2}}d\vec{L}\\=\vec{L_2}-\vec{L_1} \)

So the total angular impulse is equal to the change in angular momentum.