Question-120970

Question Number 120970 by Algoritm last updated on 04/Nov/20

Answered by mathmax by abdo last updated on 04/Nov/20

I = \int_{0}^{π} \frac{\ln (1 + \frac{cosx}{2})}{cosx} dx let f (a) = \int_{0}^{π} \frac{\ln (1 + acosx)}{cosx} dx with ∣ a ∣< 1

I = f (\frac{1}{2}) we hsve f^{^{'}} (a) = \int_{0}^{π} \frac{dx}{1 + acosx} =_{\tan (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{2 dt}{(1 + t^{2}) (1 + a \frac{1 - t^{2}}{1 + t^{2}})}

= \int_{0}^{\infty} \frac{2 dt}{1 + t^{2} + a - {at}^{2}} = \int_{0}^{\infty} \frac{2 dt}{1 + a + (1 - a) t^{2}}

= \frac{2}{1 - a} \int_{0}^{\infty} \frac{dt}{t^{2} + \frac{1 + a}{1 - a}} =_{t = \sqrt{\frac{1 + a}{1 - a}} z} \frac{2}{1 - a} . \frac{1 - a}{1 + a} \int_{0}^{\infty} \frac{1}{z^{2} + 1} \frac{\sqrt{1 + a}}{\sqrt{1 - a}} dz

= \frac{2}{\sqrt{1 - a^{2}}} \times \frac{π}{2} = \frac{π}{\sqrt{1 - a^{2}}} \Rightarrow f (a) = π \arcsin (a) + c

f (0) = 0 = 0 + c \Rightarrow c = 0 \Rightarrow f (a) = π \arcsin (a) \Rightarrow

I = f (\frac{1}{2}) = π \arcsin (\frac{1}{2}) = \frac{π^{2}}{6}