find f(x,y) =∫_0 ^(π/2) ln(x+y sinθ)dθ with ∣y∣<∣x∣ 2) find f(2,3) 3)find f((√2),(√3)) .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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 bymaxmathsup by imad last updated on 15/Oct/18
$1) we have (∂f/∂x)(x,y) =∫_0 ^(π/2) (dθ/(x+ysinθ)) changement tan((θ/2))=t give (∂f/∂x)(x,y) = ∫_0 ^1 ((2dt)/((1+t^2 )(x+y((2t)/(1+t^2 ))))) =∫_0 ^1 ((2dt)/((1+t^2 )x +2yt)) =∫_0 ^1 ((2dt)/(x +xt^2 +2yt)) =∫_0 ^1 ((2dt)/(xt^2 +2yt +x)) roots of p(x)=xt^2 +2yt +x Δ^′ =y^2 −x^2 <0 ⇒no roots but xt^2 +2yt +x =x(t^2 +2(y/x)t +1) =x( (t+(y/x))^2 +1−(y^2 /x^2 )) =x{ (t+(y/x))^2 +((x^2 −y^2 )/x^2 )} changement t+(y/x) =((√(x^2 −y^2 ))/(∣x∣)) u give ∫_0 ^1 ((2dt)/(xt^2 +2yt +x)) = ∫_(((∣x∣)/x)(y/(√(x^2 −y^2 )))) ^(((∣x∣)/x)((x+y)/(√(x^2 −y^2 )))) (2/(x((x^2 −y^2 )/x^2 ){1+u^2 }))((√(x^2 −y^2 ))/(∣x∣))du =∫_((yξ(x))/(√(x^2 −y^2 ))) ^(((x+y)ξ(x))/(√(x^2 −y^2 ))) (2/(ξ(x)(√(x^2 −y^2 )))) (du/(1+u^2 )) =(2/(ξ(x)(√(x^2 −y^2 )))) { arctan((((x+y)ξ(x))/(√(x^2 −y^2 ))))−arctan(((yξ(x))/(√(x^2 −y^2 ))))}=(∂f/∂x)(x,y) ⇒ f(x,y) =(2/(ξ(x))) ∫ (1/(√(x^2 −y^2 ))) arctan((((x+y)ξ(x))/(√(x^2 −y^2 ))))dx−(2/(ξ(x))) ∫(1/(√(x^2 −y^2 )))arctan(((yξ(x))/(√(x^2 −y^2 ))))dx+k ξ(x)=1 if x>0 and ξ(x)=−1 if x<0 ...be continued...$
$1) w e h a v e \frac{\partial f}{\partial x} (x, y) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + y s i n θ} c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e$ $\frac{\partial f}{\partial x} (x, y) = \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) (x + y \frac{2 t}{1 + t^{2}})} = \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) x + 2 y t}$ $= \int_{0}^{1} \frac{2 d t}{x + {x t}^{2} + 2 y t} = \int_{0}^{1} \frac{2 d t}{{x t}^{2} + 2 y t + x} r o o t s o f p (x) = {x t}^{2} + 2 y t + x$ $Δ^{^{'}} = y^{2} - x^{2} < 0 \Rightarrow n o r o o t s b u t {x t}^{2} + 2 y t + x = x (t^{2} + 2 \frac{y}{x} t + 1)$ $= x ({(t + \frac{y}{x})}^{2} + 1 - \frac{y^{2}}{x^{2}}) = x {{(t + \frac{y}{x})}^{2} + \frac{x^{2} - y^{2}}{x^{2}}} c h a n g e m e n t$ $t + \frac{y}{x} = \frac{\sqrt{x^{2} - y^{2}}}{∣ x ∣} u g i v e \int_{0}^{1} \frac{2 d t}{{x t}^{2} + 2 y t + x} = \int_{\frac{∣ x ∣}{x} \frac{y}{\sqrt{x^{2} - y^{2}}}}^{\frac{∣ x ∣}{x} \frac{x + y}{\sqrt{x^{2} - y^{2}}}} \frac{2}{x \frac{x^{2} - y^{2}}{x^{2}} {1 + u^{2}}} \frac{\sqrt{x^{2} - y^{2}}}{∣ x ∣} d u$ $= \int_{\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}}}^{\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}} \frac{2}{ξ (x) \sqrt{x^{2} - y^{2}}} \frac{d u}{1 + u^{2}}$ $= \frac{2}{ξ (x) \sqrt{x^{2} - y^{2}}} {a r c t a n (\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}) - a r c t a n (\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}})} = \frac{\partial f}{\partial x} (x, y) \Rightarrow$ $f (x, y) = \frac{2}{ξ (x)} \int \frac{1}{\sqrt{x^{2} - y^{2}}} a r c t a n (\frac{(x + y) ξ (x)}{\sqrt{x^{2} - y^{2}}}) d x - \frac{2}{ξ (x)} \int \frac{1}{\sqrt{x^{2} - y^{2}}} a r c t a n (\frac{y ξ (x)}{\sqrt{x^{2} - y^{2}}}) d x + k$ $ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0 . . . b e c o n t i n u e d . . .$
Commented bymaxmathsup by imad last updated on 15/Oct/18
$2)f(2,3)=∫_0 ^(π/2) ln(2+3sinθ)dθ f(2,3)= ∫_0 ^(π/2) ln(3((2/3)+sinθ))dθ =(π/2)ln(3) +∫_0 ^(π/2) ln((2/3)+sinθ)dθ let find ϕ(x)=∫_0 ^(π/2) ln(x+sinθ)dθ with ∣x∣<1 we have ϕ^′ (x) = ∫_0 ^(π/2) (dθ/(x+sinθ)) =_(tan((θ/2))=t) ∫_0 ^1 (1/(x+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^1 ((2dt)/(x(1+t^2 ) +2t)) = ∫_0 ^1 ((2dt)/(x +xt^2 +2t)) = ∫_0 ^1 ((2dt)/(xt^2 +2t +x)) Δ^′ =1−x^2 >0 ⇒t_1 =((−1+(√(1−x^2 )))/x) and t_2 =((−1−(√(1−x^2 )))/x) (we take x≠0) F(t)=(1/(xt^2 +2t +x)) =(1/(x(t−t_1 )(t−t_2 ))) =(1/x)(x/(2(√(1−x^2 )))){ (1/(t−t_1 )) −(1/(t−t_2 ))} =(1/(2(√(1−x^2 )))){ (1/(t−t_1 )) −(1/(t−t_2 ))} ⇒∫_0 ^1 F(t)dt=(1/(2(√(1−x^2 ))))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^1 =(1/(2(√(1−x^2 )))){ ln∣((1−t_1 )/(1−t_2 ))∣−ln∣(t_1 /t_2 )∣}=(1/(2(√(1−x^2 ))))ln∣((1−t_1 )/(1−t_2 )).(t_2 /t_1 )∣ =(1/(2(√(1−x^2 ))))ln∣((t_2 −t_1 t_2 )/(t_1 −t_1 t_2 ))∣ =(1/(2(√(1−x^2 ))))ln∣ ((((−1−(√(1−x^2 )))/x) −1)/(((−1+(√(1−x^2 )))/x)−1))∣ =(1/(2(√(1−x^2 ))))ln∣ ((1+((1+(√(1−x^2 )))/x))/(1−((−1+(√(1−x^2 )))/x)))∣=(1/(2(√(1−x^2 ))))ln∣ ((x+1+(√(1−x^2 )))/(x+1−(√(1−x^2 ))))∣ ⇒ ϕ^′ (x) =(1/(√(1−x^2 )))ln(((x+1+(√(1−x^2 )))/(x+1−(√(1−x^2 ))))) ⇒ ϕ(x) =∫_0 ^x (1/(√(1−u^2 )))ln(((u+1+(√(1−u^2 )))/(u+1−(√(1−u^2 )))))du +c c=ϕ(0)=∫_0 ^(π/2) ln(sinθ)dθ =−(π/2)ln(2) ⇒ ϕ(x) =∫_0 ^x (1/(√(1−u^2 )))ln(((u+1+(√(1−u^2 )))/(u+1−(√(1−u^2 )))))du−(π/2)ln(2) ...be continued...$
$2) f (2, 3) = \int_{0}^{\frac{π}{2}} l n (2 + 3 s i n θ) d θ$ $f (2, 3) = \int_{0}^{\frac{π}{2}} l n (3 (\frac{2}{3} + s i n θ)) d θ = \frac{π}{2} l n (3) + \int_{0}^{\frac{π}{2}} l n (\frac{2}{3} + s i n θ) d θ$ $l e t f i n d φ (x) = \int_{0}^{\frac{π}{2}} l n (x + s i n θ) d θ w i t h ∣ x ∣< 1 w e h a v e$ $φ^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + s i n θ} =_{t a n (\frac{θ}{2}) = t} \int_{0}^{1} \frac{1}{x + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{1} \frac{2 d t}{x (1 + t^{2}) + 2 t}$ $= \int_{0}^{1} \frac{2 d t}{x + {x t}^{2} + 2 t} = \int_{0}^{1} \frac{2 d t}{{x t}^{2} + 2 t + x}$ $Δ^{^{'}} = 1 - x^{2} > 0 \Rightarrow t_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x} a n d t_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x} (w e t a k e x \neq 0)$ $F (t) = \frac{1}{{x t}^{2} + 2 t + x} = \frac{1}{x (t - t_{1}) (t - t_{2})} = \frac{1}{x} \frac{x}{2 \sqrt{1 - x^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}}$ $= \frac{1}{2 \sqrt{1 - x^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} \Rightarrow \int_{0}^{1} F (t) d t = \frac{1}{2 \sqrt{1 - x^{2}}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{1}$ $= \frac{1}{2 \sqrt{1 - x^{2}}} {l n ∣ \frac{1 - t_{1}}{1 - t_{2}} ∣ - l n ∣ \frac{t_{1}}{t_{2}} ∣} = \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{1 - t_{1}}{1 - t_{2}} . \frac{t_{2}}{t_{1}} ∣$ $= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{t_{2} - t_{1} t_{2}}{t_{1} - t_{1} t_{2}} ∣ = \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{\frac{- 1 - \sqrt{1 - x^{2}}}{x} - 1}{\frac{- 1 + \sqrt{1 - x^{2}}}{x} - 1} ∣$ $= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{1 + \frac{1 + \sqrt{1 - x^{2}}}{x}}{1 - \frac{- 1 + \sqrt{1 - x^{2}}}{x}} ∣= \frac{1}{2 \sqrt{1 - x^{2}}} l n ∣ \frac{x + 1 + \sqrt{1 - x^{2}}}{x + 1 - \sqrt{1 - x^{2}}} ∣ \Rightarrow$ $φ^{^{'}} (x) = \frac{1}{\sqrt{1 - x^{2}}} l n (\frac{x + 1 + \sqrt{1 - x^{2}}}{x + 1 - \sqrt{1 - x^{2}}}) \Rightarrow$ $φ (x) = \int_{0}^{x} \frac{1}{\sqrt{1 - u^{2}}} l n (\frac{u + 1 + \sqrt{1 - u^{2}}}{u + 1 - \sqrt{1 - u^{2}}}) d u + c$ $c = φ (0) = \int_{0}^{\frac{π}{2}} l n (s i n θ) d θ = - \frac{π}{2} l n (2) \Rightarrow$ $φ (x) = \int_{0}^{x} \frac{1}{\sqrt{1 - u^{2}}} l n (\frac{u + 1 + \sqrt{1 - u^{2}}}{u + 1 - \sqrt{1 - u^{2}}}) d u - \frac{π}{2} l n (2) . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum