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Question Number 95221 by mathmax by abdo last updated on 24/May/20

calculate \int_{0}^{π} \ln (2 + \cos θ) d θ

Answered by mathmax by abdo last updated on 24/May/20

I = \int_{0}^{π} \ln (2 + \cos θ) d θ \Rightarrow I = π \ln (2) + \int_{0}^{π} \ln (1 + \frac{1}{2} \cos θ)

let introduce the parametric function f (a) = \int_{0}^{π} \ln (1 + acos θ) d θ

with 0 < a < 1 we have f^{^{'}} (a) = \int_{0}^{π} \frac{\cos θ}{1 + acos θ} d θ

= \frac{1}{a} \int_{0}^{π} \frac{1 + acos θ - 1}{1 + acos θ} d θ = \frac{π}{a} - \frac{1}{a} \int_{0}^{π} \frac{d θ}{1 + acos θ} we have

\int_{0}^{π} \frac{d θ}{1 + acos θ} =_{\tan (\frac{θ}{2}) = t} \int_{0}^{\infty} \frac{2 dt}{(1 + t^{2}) (1 + a \frac{1 - t^{2}}{1 + t^{2}})}

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

=_{t = \sqrt{\frac{1 + a}{1 - a}} u} \frac{2}{1 - a} \times \frac{1 - a}{1 + a} \int_{0}^{\infty} \frac{1}{u^{2} + 1} \times \sqrt{\frac{1 + a}{1 - a}} du

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

f (a) = π \ln (a) - π arcsina + c

f (1) = - \frac{π^{2}}{2} + c \Rightarrow c = \frac{π^{2}}{2} + \int_{0}^{π} \ln (1 + \cos θ) d θ but

\int_{0}^{π} \ln (1 + \cos θ) d θ = \int_{0}^{π} \ln ({2 \cos}^{2} (\frac{θ}{2})) d θ

= π \ln (2) + 2 \int_{0}^{π} \ln (\cos (\frac{θ}{2})) d θ (\frac{θ}{2} = z)

= π \ln (2) + 2 \int_{0}^{\frac{π}{2}} \ln (cosz) dz = π \ln (2) + 2 (- \frac{π}{2} \ln (2)) = 0 \Rightarrow c = \frac{π^{2}}{2}

\Rightarrow f (a) = π lna - π arcsina + \frac{π^{2}}{2} and

I = π \ln (2) + f (\frac{1}{2}) = π \ln (2) - π \ln (2) - π \arcsin (\frac{1}{2}) + \frac{π^{2}}{2}

= \frac{π^{2}}{2} - π (\frac{π}{6}) = \frac{π^{2}}{2} - \frac{π^{2}}{6} = \frac{2 π^{2}}{6} = \frac{π^{2}}{3} \Rightarrow

\int_{0}^{π} \ln (2 + \cos θ) d θ = \frac{π^{2}}{3}