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Question Number 45600 by maxmathsup by imad last updated on 14/Oct/18

f i n d f (x, y) = \int_{0}^{\frac{π}{2}} l n (x + y s i n θ) d θ w i t h ∣ y ∣<∣ x ∣

2) f i n d f (2, 3)

3) f i n d f (\sqrt{2}, \sqrt{3}) .

Commented by maxmathsup by imad last updated on 15/Oct/18

1) w e h a v e \frac{\partial f}{\partial x} (x, y) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + y s i n θ} c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e

\frac{\partial f}{\partial x} (x, y) = \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) (x + y \frac{2 t}{1 + t^{2}})} = \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) x + 2 y t}

= \int_{0}^{1} \frac{2 d t}{x + {x t}^{2} + 2 y t} = \int_{0}^{1} \frac{2 d t}{{x t}^{2} + 2 y t + x} r o o t s o f p (x) = {x t}^{2} + 2 y t + x

Δ^{^{'}} = y^{2} - x^{2} < 0 \Rightarrow n o r o o t s b u t {x t}^{2} + 2 y t + x = x (t^{2} + 2 \frac{y}{x} t + 1)

= x ({(t + \frac{y}{x})}^{2} + 1 - \frac{y^{2}}{x^{2}}) = x {{(t + \frac{y}{x})}^{2} + \frac{x^{2} - y^{2}}{x^{2}}} c h a n g e m e n t

t + \frac{y}{x} = \frac{\sqrt{x^{2} - y^{2}}}{∣ x ∣} u g i v e \int_{0}^{1} \frac{2 d t}{{x t}^{2} + 2 y t + x} = \int_{\frac{∣ x ∣}{x} \frac{y}{\sqrt{x^{2} - y^{2}}}}^{\frac{∣ x ∣}{x} \frac{x + y}{\sqrt{x^{2} - y^{2}}}} \frac{2}{x \frac{x^{2} - y^{2}}{x^{2}} {1 + u^{2}}} \frac{\sqrt{x^{2} - y^{2}}}{∣ x ∣} d u

= \int_{\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}}}^{\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}} \frac{2}{ξ (x) \sqrt{x^{2} - y^{2}}} \frac{d u}{1 + u^{2}}

= \frac{2}{ξ (x) \sqrt{x^{2} - y^{2}}} {a r c t a n (\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}) - a r c t a n (\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}})} = \frac{\partial f}{\partial x} (x, y) \Rightarrow

f (x, y) = \frac{2}{ξ (x)} \int \frac{1}{\sqrt{x^{2} - y^{2}}} a r c t a n (\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}) d x - \frac{2}{ξ (x)} \int \frac{1}{\sqrt{x^{2} - y^{2}}} a r c t a n (\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}}) d x + k

ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0 \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 15/Oct/18

2) f (2, 3) = \int_{0}^{\frac{π}{2}} l n (2 + 3 s i n θ) d θ

f (2, 3) = \int_{0}^{\frac{π}{2}} l n (3 (\frac{2}{3} + s i n θ)) d θ = \frac{π}{2} l n (3) + \int_{0}^{\frac{π}{2}} l n (\frac{2}{3} + s i n θ) d θ

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

φ^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + s i n θ} =_{t a n (\frac{θ}{2}) = t} \int_{0}^{1} \frac{1}{x + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{1} \frac{2 d t}{x (1 + t^{2}) + 2 t}

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

Δ^{^{'}} = 1 - x^{2} > 0 \Rightarrow t_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x} a n d t_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x} (w e t a k e x \neq 0)

F (t) = \frac{1}{{x t}^{2} + 2 t + x} = \frac{1}{x (t - t_{1}) (t - t_{2})} = \frac{1}{x} \frac{x}{2 \sqrt{1 - x^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}}

= \frac{1}{2 \sqrt{1 - x^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} \Rightarrow \int_{0}^{1} F (t) d t = \frac{1}{2 \sqrt{1 - x^{2}}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{1}

= \frac{1}{2 \sqrt{1 - x^{2}}} {l n ∣ \frac{1 - t_{1}}{1 - t_{2}} ∣ - l n ∣ \frac{t_{1}}{t_{2}} ∣} = \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{1 - t_{1}}{1 - t_{2}} . \frac{t_{2}}{t_{1}} ∣

= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{t_{2} - t_{1} t_{2}}{t_{1} - t_{1} t_{2}} ∣ = \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{\frac{- 1 - \sqrt{1 - x^{2}}}{x} - 1}{\frac{- 1 + \sqrt{1 - x^{2}}}{x} - 1} ∣

= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{1 + \frac{1 + \sqrt{1 - x^{2}}}{x}}{1 - \frac{- 1 + \sqrt{1 - x^{2}}}{x}} ∣= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{x + 1 + \sqrt{1 - x^{2}}}{x + 1 - \sqrt{1 - x^{2}}} ∣ \Rightarrow

φ^{^{'}} (x) = \frac{1}{\sqrt{1 - x^{2}}} l n (\frac{x + 1 + \sqrt{1 - x^{2}}}{x + 1 - \sqrt{1 - x^{2}}}) \Rightarrow

φ (x) = \int_{0}^{x} \frac{1}{\sqrt{1 - u^{2}}} l n (\frac{u + 1 + \sqrt{1 - u^{2}}}{u + 1 - \sqrt{1 - u^{2}}}) d u + c

c = φ (0) = \int_{0}^{\frac{π}{2}} l n (s i n θ) d θ = - \frac{π}{2} l n (2) \Rightarrow

φ (x) = \int_{0}^{x} \frac{1}{\sqrt{1 - u^{2}}} l n (\frac{u + 1 + \sqrt{1 - u^{2}}}{u + 1 - \sqrt{1 - u^{2}}}) d u - \frac{π}{2} l n (2) \dots b e c o n t i n u e d \dots