4 ms·
I don't really understand the part from how did the author jumps from -pi*(1+a^2)/(1-a^2) to df/da = 2pi/a. Anyone knows how the author did it?
by Trung0246 5y ago
I don't really understand the part from how did the author jumps from -pi*(1+a^2)/(1-a^2) to df/da = 2pi/a. Anyone knows how the author did it?
- quibono 5y agoNotice 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