A circle of radius $\mathrm{a}$ has charge density given by $\mathrm{\lambda }={\mathrm{\lambda }}_0{\cos }^2\mathrm{\theta }$ on its circumference. What will be the total charge on the circle ?

$\mathrm{\lambda }={\mathrm{\lambda }}_0{\cos }^2\mathrm{\theta }={\mathrm{\lambda }}_0\left(\frac{1+\cos 2\mathrm{\theta }}{2}\right)$

$\Rightarrow \mathrm{\lambda }=\frac{{\mathrm{\lambda }}_0}{2}+\frac{{\mathrm{\lambda }}_0}{2}\cos 2\mathrm{\theta }$

$\mathrm{dq}=\mathrm{\lambda rd\theta }$

$\Rightarrow \mathrm{dq}=\left(\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}+\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}\cos 2\mathrm{\theta }\right)\mathrm{d\theta }$

$\Rightarrow \mathrm{q}=\int \mathrm{dq}={\int }_0^{2\mathrm{\pi }}\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}\mathrm{d\theta }+{\int }_0^{2\mathrm{\pi }}\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}\cos 2\mathrm{\theta }\cdot \mathrm{d\theta }$

${\left.=\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}\left(2\mathrm{\pi }\right)+\frac{{\mathrm{\lambda }}_0\mathrm{a}}{2}\left(\frac{\sin 2\mathrm{\theta }}{2}\right)\right]}_0^{2\mathrm{\pi }}={\mathrm{\lambda }}_0\mathrm{\pi a}$