1-find-g-x-0-pi-2-ln-1-x-2-cos-2-d-with-x-from-R-2-find-the-value-of-0-pi-2-ln-1-2-cos-2-d-and-3-find-the-value-of-A-0-pi-2-ln-1-cos-2-cos-2-d-

Question Number 40661 by math khazana by abdo last updated on 25/Jul/18

1) f i n d g (x) = \int_{0}^{\frac{π}{2}} l n (1 - x^{2} {c o s}^{2} θ) d θ w i t h x f r o m R

2) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} l n (1 - 2 {c o s}^{2} θ) d θ a n d

3) f i n d t h e v a l u e o f

A (α) = \int_{0}^{\frac{π}{2}} l n (1 - {c o s}^{2} α {c o s}^{2} θ) d θ

Commented by math khazana by abdo last updated on 04/Aug/18

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

+ \int_{0}^{\frac{π}{2}} l n (1 - x c o s θ) d θ = h (x) + k (x)

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

h^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{c o s θ}{1 + x c o s θ} d θ

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

b u t \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + x c o s θ} =_{t a n (\frac{θ}{2}) = t} \int_{0}^{1} \frac{1}{1 + x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \frac{2}{(1 + x)} \int_{0}^{1} \frac{d t}{1 + \frac{1 - x}{1 + x} t^{2}} i f ∣ x ∣< 1 w e d o t h e

c h a n g e m e n t \sqrt{\frac{1 - x}{1 + x}} t = u \Rightarrow

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

= \frac{2}{\sqrt{1 - x^{2}}} a r c t a n \sqrt{\frac{1 - x}{1 + x}} \Rightarrow

h (x) = \int \frac{2 a r c t a n \sqrt{\frac{1 - x}{1 + x}}}{\sqrt{1 - x^{2}}} d x + c

c h a n g e m e n t x = c o s t g i v e

h (x) = \int \frac{2 a r c t a n (t a n (\frac{t}{2}))}{s i n t} (- s i n t) d t + c

= - \int t d t + c = - \frac{t^{2}}{2} + c = c - \frac{1}{2} {(a r c c o s x)}^{2}

h (0) = 0 = c - \frac{1}{2} \Rightarrow c = \frac{1}{2} \Rightarrow

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

Commented by math khazana by abdo last updated on 04/Aug/18

e r r o r f r o m l i n e 12 w e h a v e

h^{^{'}} (x) = \frac{π}{2 x} - \frac{1}{x} \frac{2}{\sqrt{1 - x^{2}}} a r c t a n \sqrt{\frac{1 - x}{1 + x}} \Rightarrow

h (x) = \frac{π}{2} l n ∣ x ∣ - \int \frac{2 a r c t a n \sqrt{\frac{1 - x}{1 + x}}}{x \sqrt{1 - x^{2}}} d x + c b u t

c h a n g e m e n t x = c o s t g i v e

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

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

= - \int \frac{a r c t a n u}{1 - u^{2}} d u a n y w a y w e h a v e

h (x) = \frac{π}{2} l n ∣ x ∣ + \int_{1}^{x} \frac{t}{c o s t} d t + c

c = h (1) = \int_{0}^{\frac{π}{2}} l n (1 + c o s θ) d θ \Rightarrow

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

i f ∣ x ∣< 1

i f ∣ x ∣> 1 w e f o l l o w t h e s a m e w a y

Commented by math khazana by abdo last updated on 04/Aug/18

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

w e h a v e g (x) = h (x) + k (x) \Rightarrow k (x) = g (x) - h (x) \Rightarrow

k (- x) = g (- x) - h (- x) = g (x) - h (x) = k (x)

b e c a u s e g a n d h a r e e v e n s o

k (x) = k (- x) = h (x) \Rightarrow

g (x) = 2 h (x)

Commented by math khazana by abdo last updated on 04/Aug/18

g (x) = π l n ∣ x ∣ + 2 \int_{1}^{x} \frac{t}{c o s t} d t + 2 \int_{0}^{\frac{π}{2}} l n (1 + c o s θ) d θ .