For radiance in every incoming d_omega, only the part of radiance that is perpendicular to the surface effectively contributes to the irradiance on point p. That's why it needs to multiply by cos(theta). Surprisingly the integral of all perpendicular radiance is exactly half of the integral of radiance over the hemisphere.
For radiance in every incoming d_omega, only the part of radiance that is perpendicular to the surface effectively contributes to the irradiance on point p. That's why it needs to multiply by cos(theta). Surprisingly the integral of all perpendicular radiance is exactly half of the integral of radiance over the hemisphere.