3 ms·
Notice that it is actually df/da = (pi/a) - (1/a) (1-a^2)/(1+a^2) int_0^pi (1 - (1-a^2) / (1 + a^2 + 2 a cosx) dx. It is that right integral that is equal to -
by quibono 5y ago
Notice that it is actually df/da = (pi/a) - (1/a) (1-a^2)/(1+a^2) int_0^pi (1 - (1-a^2) / (1 + a^2 + 2 a cosx) dx.
It is that right integral that is equal to -pi (1+a^2)/(1-a^2).
So when you add you get:
pi/a - 1/a (1-a^2)/(1+a^2) * (-pi (1+a^2)/(1-a^2)) =
pi /a - (-pi/a) = 2 pi/a