find-0-pi-ln-2-cost-dt-and-0-pi-ln-2-cost-dt-

Question Number 38899 by math khazana by abdo last updated on 01/Jul/18

f i n d \int_{0}^{π} l n (2 + c o s t) d t a n d \int_{0}^{π} l n (2 - c o s t) d t

Commented by maxmathsup by imad last updated on 11/Jul/18

l e t I = \int_{0}^{π} l n (2 + c o s t) d t

I = π l n (2) + \int_{0}^{π} l n (1 + \frac{1}{2} c o s t) t l e t n t r o d u c e t h e p a r a m e t r i c f u n c t i o n

f (x) = \int_{0}^{π} l n (1 + x c o s t) d t w e h a v e I = f (\frac{1}{2}) (w e t a k e ∣ x ∣< 1)

f^{^{'}} (x) = \int_{0}^{π} \frac{c o s t}{1 + x c o s t} d t = \frac{1}{x} \int_{0}^{π} \frac{1 + x c o s t - 1}{1 + x c o s t} d t

= \frac{π}{x} - \frac{1}{x} \int_{0}^{π} \frac{d t}{1 + x c o s t} d t b u t c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e

\int_{0}^{π} \frac{d t}{1 + x c o s t} = \int_{0}^{\infty} \frac{1}{1 + x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{\infty} \frac{2 d u}{1 + u^{2} + x (1 - u^{2})} = \int_{0}^{\infty} \frac{2 d u}{1 + x + (1 - x) u^{2}}

= \frac{2}{1 + x} \int_{0}^{\infty} \frac{d u}{1 + \frac{1 - x}{1 + x} u^{2}} =_{\sqrt{\frac{1 - x}{1 + x} u} = α} \frac{2}{1 + x} \int_{0}^{\infty} \frac{1}{1 + α^{2}} \sqrt{\frac{1 + x}{1 - x}} d α

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

f (x) = π l n ∣ x ∣ - π \int_{} \frac{d x}{x \sqrt{1 - x^{2}}} + c

\int \frac{d x}{x \sqrt{1 - x^{2}}} =_{x = s i n θ} \int \frac{c o s θ d θ}{s i n θ c o s θ} = \int \frac{d θ}{s i n θ} =_{t a n (\frac{θ}{2}) = u} \int \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int \frac{d u}{u} = l n ∣ u ∣ = l n ∣ t a n (\frac{θ}{2}) ∣ = l n ∣ t a n (\frac{a r c i n x}{2}) ∣ \Rightarrow

f (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{a r c s i n x}{2}) ∣ + c

w e h a v e f (1) = c = \int_{0}^{π} l n (1 + c o s x) d x = \int_{0}^{π} l n (2 {c o s}^{2} (\frac{x}{2})) d x

= π l n (2) + 2 \int_{0}^{π} l n (c o s (\frac{x}{2})) d x

=_{\frac{x}{2} = t} π l n (2) + 2 \int_{0}^{\frac{π}{2}} l n (c o s t) (2 d t) = π l n (2) + 4 (- \frac{π}{2} l n (2))

= - π l n (2) \Rightarrow f (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{a r c s i n x}{2}) ∣ - π l n (2)

I = f (\frac{1}{2}) = - 2 π l n (2) - π l n ∣ t a n (\frac{π}{12}) ∣

{c o s}^{2} (\frac{π}{12}) = \frac{1 + c o s (\frac{π}{6})}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} \Rightarrow c o s (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}

{s i n}^{2} (\frac{π}{12}) = \frac{1 - c o s (\frac{π}{6})}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4} \Rightarrow s i n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}

t a n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} = \frac{2 - \sqrt{3}}{\sqrt{4 - 3}} = 2 - \sqrt{3} . \Rightarrow

I = - 2 π l n (2) - π l n (2 - \sqrt{3})

Commented by maxmathsup by imad last updated on 11/Jul/18

l e t J = \int_{0}^{π} l n (2 - c o s t) d t

J = π l n (2) + \int_{0}^{π} l n (1 - \frac{1}{2} c o s t) d t = π l n (2) + f (- \frac{1}{2})

= π l n (2) + f (\frac{1}{2}) (f i s o d d)

= π l n (2) - 2 π l n (2) - π l n (2 - \sqrt{3}) = - π l n (2) - π l n (2 - \sqrt{3})

Commented by maxmathsup by imad last updated on 11/Jul/18

I = π l n (2) + f (\frac{1}{2}) = - π l n (2) - π l n (2 - \sqrt{3}) .

Commented by maxmathsup by imad last updated on 11/Jul/18

f i s e v e n .