If an external torque \( \tau \) acts on a particle during the time from \( t_1 \) to \( t_2 \), then the angular impulse is given by,

\( \displaystyle{\int_{t_1}^{t_2}}\vec{\tau}\ dt \)

Again we know that the torque acting on a particle is equal to the time rate of change of linear momentum of that particle, So

\( \vec{\tau}=\frac{d\vec{L}}{dt} \)
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Let us consider from time \( t_1 \) to \( t_2 \), the angular momentum changes from \( \vec{L_1} \) to \( \vec{L_2} \). So we can write,

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

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