1-find-f-x-0-pi-ln-2-x-cos-d-2-calculate-0-pi-ln-2-cos-d-

Question Number 38718 by maxmathsup by imad last updated on 28/Jun/18

1) f i n d f (x) = \int_{0}^{π} l n (2 + x c o s θ) d θ

2) c a l c u l a t e \int_{0}^{π} l n (2 + c o s θ) d θ

Commented by math khazana by abdo last updated on 30/Jun/18

1) w e h a v e f^{^{'}} (x) = \int_{0}^{π} \frac{c o s θ}{2 + x c o s θ} d θ c h a n g e m e n t

t a n (\frac{θ}{2}) = t \Rightarrow f^{^{'}} (x) = \int_{0}^{\infty} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{2 + x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

f^{^{'}} (x) = 2 \int_{0}^{\infty} \frac{1 - t^{2}}{(1 + t^{2}) (2 + 2 t^{2} + x - {x t}^{2})} d t

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

f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{1 - t^{2}}{(1 + t^{2}) {(2 - x) t^{2} + x + 2}} d t l e t

φ (z) = \frac{1 - z^{2}}{(z^{2} + 1) {(2 - x) z^{2} + x + 2}} i f x \neq 2

φ (z) = \frac{1 - z^{2}}{(2 - x) (z^{2} + 1) {z^{2} + \frac{2 + x}{2 - x}}}

c a s e 1 \frac{2 + x}{2 - x} > 0 \Rightarrow

φ (z) = \frac{1 - z^{2}}{(2 - x) (z - i) (z + i) (z - i \sqrt{\frac{2 + x}{2 - x}}) (z + i \sqrt{\frac{2 + x}{2 - x}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, i \sqrt{\frac{2 + x}{2 - x}})}

R e s (φ, i) = \frac{2}{(2 - x) (2 i) (- 1 + \frac{2 + x}{2 - x})}

= \frac{- i (2 - x)}{(2 - x) (- 2 + x + 2 + x)} = \frac{- i}{2 x}

R e s (φ, i \sqrt{\frac{2 + x}{2 - x}}) = \frac{1 + \frac{2 + x}{2 - x}}{(2 - x) (- \frac{2 + x}{2 - x} + 1) 2 i \sqrt{\frac{2 + x}{2 - x}}}

= \frac{4}{{(2 - x)}^{2} (\frac{- 2 - x + 2 - x}{2 - x}) 2 i \sqrt{\frac{2 + x}{2 - x}}}

= \frac{- 2 i}{(2 - x) (- 2 x) \frac{\sqrt{2 + x}}{\sqrt{2 - x}}} = \frac{i}{\sqrt{4 - x^{2}}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{- i}{2 x} + \frac{i}{\sqrt{4 - x^{2}}}}

= \frac{π}{x} + \frac{2 π}{\sqrt{4 - x^{2}}} = f^{^{'}} (x) \Rightarrow

f (x) = π l n ∣ x ∣ + 2 π \int \frac{d x}{\sqrt{4 - x^{2}}} + c c h a n g e m e n t

x = 2 u g i v e \int \frac{d x}{\sqrt{4 - x^{2}}} = \int \frac{2 d u}{2 \sqrt{1 - u^{2}}}

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

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

f (x) = π l n ∣ x ∣ + 2 π a r c s i n (\frac{x}{2}) + f (1) - π^{2} .

Commented by math khazana by abdo last updated on 30/Jun/18

c a s e 2 \frac{2 + x}{2 - x} < 0 w e f o l l o w t h e s a m e m e t h o d . .

Commented by math khazana by abdo last updated on 30/Jun/18

e r r o r f r o m l i n e 15

R e s (φ, i \sqrt{\frac{2 + x}{2 - x}}) = \frac{i}{x \sqrt{4 - x^{2}}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{- i}{2 x} + \frac{i}{x \sqrt{4 - x^{2}}}}

= \frac{π}{x} - \frac{2 π}{x \sqrt{4 - x^{2}}} = f^{^{'}} (x) \Rightarrow

f (x) = π l n ∣ x ∣ - 2 π \int \frac{d x}{x \sqrt{4 - x^{2}}} . c h a n g e m e n t

x = 2 s i n t \Rightarrow

\int \frac{d x}{x \sqrt{4 - x^{2}}} = \int \frac{2 c o s t}{2 s i n t 2 c o s t} d t

= \frac{1}{2} \int \frac{d t}{s i n t} t h e n c h a n g . t a n (\frac{t}{2}) = u

= \frac{1}{2} \int \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} l n ∣ t a n (\frac{t}{2}))

= \frac{1}{2} l n ∣ t a n (\frac{1}{2} a r c s i n (\frac{x}{2})) ∣\Rightarrow

f (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{1}{2} a r c s i n (\frac{x}{2}) ∣ + c

f (1) = c \Rightarrow

f (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{1}{2} a r c s i n (\frac{x}{2}) ∣ + \int_{0}^{π} l n (2 + c o s θ) d θ