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Question Number 103345 by harckinwunmy last updated on 14/Jul/20

Commented by harckinwunmy last updated on 14/Jul/20

$please help$
Commented by Dwaipayan Shikari last updated on 14/Jul/20
$2∫_0 ^0 (((2t)/(1+t^2 ))/((5/4)−((1−t^2 )/(1+t^2 )))) .(1/(1+t^2 ))dt=0 {take tan(θ/2)=t$
$2 \int_{0}^{0} \frac{\frac{2 t}{1 + t^{2}}}{\frac{5}{4} - \frac{1 - t^{2}}{1 + t^{2}}} . \frac{1}{1 + t^{2}} d t = 0 {t a k e t a n \frac{θ}{2} = t$
Commented by mathmax by abdo last updated on 14/Jul/20

$not correct sir you must have a bijection . . .!$
Commented by Dwaipayan Shikari last updated on 14/Jul/20

$\int_{0}^{0} f (x) d x = 0$
Commented by mathmax by abdo last updated on 15/Jul/20

$\tan is bijective on] - \frac{π}{2}, \frac{π}{2} [not [0, 2 π] be carefull$
	Answered by abdomathmax last updated on 14/Jul/20

	$Q 2 \frac{\partial u}{\partial x} (x, y) = \frac{1 (x^{2} + y^{2}) - 2 x (x - y)}{{(x^{2} + y^{2})}^{2}}$ $= \frac{- x^{2} + y^{2} + 2 xy}{{(x^{2} + y^{2})}^{2}}$ $\frac{\partial^{2} u}{\partial x^{2}} (x, y) = \frac{(- 2 x + 2 y) {(x^{2} + y^{2})}^{2} - 2 (2 x) (x^{2} + y^{2})}{{(x^{2} + y^{2})}^{4}}$ $= \frac{(- 2 x + 2 y) (x^{2} + y^{2}) - 4 x}{{(x^{2} + y^{2})}^{3}}$ $= \frac{- {2 x}^{3} - {2 xy}^{2} + {2 yx}^{2} - 4 x}{{(x^{2} + y^{2})}^{3}} after we calculate$ $\frac{\partial^{2} u}{\partial y^{2}} (x, y) and prove that \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0$
	Answered by abdomathmax last updated on 14/Jul/20

	$let I = \int_{0}^{5} {(25 - t^{2})}^{\frac{3}{2}} dt we do the changeme 7 t$ $t = 5 x \Rightarrow I = \int_{0}^{1} {(25 - {25 x}^{2})}^{\frac{3}{2}} 5 dx$ $= 5 . {(5^{2})}^{\frac{3}{2}} \int_{0}^{1} {(1 - x^{2})}^{\frac{3}{2}} dx$ $=_{x = u^{\frac{1}{2}}} 5^{4} \int_{0}^{1} {(1 - u)}^{\frac{3}{2}} \frac{1}{2} u^{- \frac{1}{2}} du$ $= \frac{5^{4}}{2} \int_{0}^{1} u^{- \frac{1}{2}} {(1 - u)}^{\frac{3}{2}} du we havd$ $B (p, q) = \int_{0}^{1} x^{p - 1} {(1 - x)}^{q - 1} dx$ $p - 1 = - \frac{1}{2} \Rightarrow p = \frac{1}{2} and q - 1 = \frac{3}{2} \Rightarrow q = 1 + \frac{3}{2} = \frac{5}{2}$ $\Rightarrow I = \frac{5^{4}}{2} \times B (\frac{1}{2}, \frac{5}{2}) .$
	Answered by abdomathmax last updated on 14/Jul/20
	$A =∫_0 ^(2π) ((sinθ)/((5/4)−cosθ))dθ ⇒A =4∫_0 ^(2π ) ((sinθ)/(5−4cosθ))dθ changement e^(iθ) =z give A =4 ∫_(∣z∣=1) (((z−z^(−1) )/(2i))/(5−4×((z+z^(−1) )/2)))×(dz/(iz)) =−2 ∫_(∣z∣=1) ((2(z−z^(−1) ))/(z(10−4z−4z^(−1) )))dz =−4 ∫_(∣z∣=1) ((z−z^(−1) )/(10z−4z^2 −4z))dz =4 ∫_(∣z∣=1) ((z^2 −1)/(z(4z^2 −10z +4)))dz =2 ∫_(∣z∣=1) ((z^2 −1)/(z(2z^2 −5z +2)))dz let ϕ(z) =((z^2 −1)/(z{2z^2 −5z +2})) poles of ϕ? 2z^2 −5z +2 =0 →Δ =25−16 =9 ⇒ z_1 =((5+3)/4) =2 and z_2 =((5−3)/4) =(1/2) ∫_(∣z∣ =1) ϕ(z)dz =2iπ {Res(ϕ,0) +Res(ϕ,z_2 )} ϕ(z) =((z^2 −1)/(2z(z−2)(z−(1/2)))) Res(ϕ,0) =lim_(z→0) zϕ(z) =−(1/2) Res(ϕ,(1/2)) =lim_(z→(1/2)) (z−(1/2))ϕ(z) =(((1/4)−1)/(2×(1/2)((1/2)−2))) =−(3/4)×(1/(−(3/2))) =(1/2) ⇒ ∫_(∣z∣=1) ϕ(z)dz =0 ⇒ A =0$
	$A = \int_{0}^{2 π} \frac{\sin θ}{\frac{5}{4} - \cos θ} d θ \Rightarrow A = 4 \int_{0}^{2 π} \frac{\sin θ}{5 - 4 \cos θ} d θ$ $changement e^{i θ} = z give$ $A = 4 \int_{∣ z ∣= 1} \frac{\frac{z - z^{- 1}}{2 i}}{5 - 4 \times \frac{z + z^{- 1}}{2}} \times \frac{dz}{iz}$ $= - 2 \int_{∣ z ∣= 1} \frac{2 (z - z^{- 1})}{z (10 - 4 z - {4 z}^{- 1})} dz$ $= - 4 \int_{∣ z ∣= 1} \frac{z - z^{- 1}}{10 z - {4 z}^{2} - 4 z} dz$ $= 4 \int_{∣ z ∣= 1} \frac{z^{2} - 1}{z ({4 z}^{2} - 10 z + 4)} dz$ $= 2 \int_{∣ z ∣= 1} \frac{z^{2} - 1}{z ({2 z}^{2} - 5 z + 2)} dz$ $let φ (z) = \frac{z^{2} - 1}{z {{2 z}^{2} - 5 z + 2}} poles of φ ?$ ${2 z}^{2} - 5 z + 2 = 0 \to Δ = 25 - 16 = 9 \Rightarrow$ $z_{1} = \frac{5 + 3}{4} = 2 and z_{2} = \frac{5 - 3}{4} = \frac{1}{2}$ $\int_{∣ z ∣ = 1} φ (z) dz = 2 i π {Res (φ, 0) + Res (φ, z_{2})}$ $φ (z) = \frac{z^{2} - 1}{2 z (z - 2) (z - \frac{1}{2})}$ $Res (φ, 0) = \lim_{z \to 0} z φ (z) = - \frac{1}{2}$ $Res (φ, \frac{1}{2}) = \lim_{z \to \frac{1}{2}} (z - \frac{1}{2}) φ (z)$ $= \frac{\frac{1}{4} - 1}{2 \times \frac{1}{2} (\frac{1}{2} - 2)} = - \frac{3}{4} \times \frac{1}{- \frac{3}{2}} = \frac{1}{2} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) dz = 0 \Rightarrow A = 0$
	Answered by 1549442205 last updated on 15/Jul/20

	$Q 3 . F = \int \frac{\sin θ}{\frac{5}{4} - \cos θ} d θ = - \int \frac{dcos θ}{\frac{5}{4} - \cos θ} \underset{u = \cos θ}{=} \int \frac{du}{u - \frac{5}{4}}$ $. \int \frac{d (u - \frac{5}{4})}{u - \frac{5}{4}} = \ln ∣ u - \frac{5}{4} ∣= \ln (\frac{5}{4} - \cos θ) ∣_{0}^{2 π}$ $= \ln (\frac{1}{4}) - \ln (\frac{1}{4}) = 0$ $Q 2 . b / \int_{0}^{5} {(25 - t^{2})}^{\frac{3}{2}} dt = F . Put t = 5 \sin θ$ $\Rightarrow dt = 5 \cos θ d θ . Then$ $F = \int_{0}^{\frac{π}{2}} {({25 \cos}^{2} θ)}^{\frac{3}{2}} 5 \cos θ d θ = 625 \int_{0}^{\frac{π}{2}} {(\cos θ)}^{4} d θ$ $= \frac{625}{2} \int_{0}^{\frac{π}{2}} 2 {(\sin θ)}^{2 . \frac{1}{2} - 1} {(\cos θ)}^{2 . \frac{5}{2} - 1} d θ$ $= \frac{625}{2} B (\frac{5}{2}, \frac{1}{2}) = \frac{625}{2} \times \frac{Γ (\frac{5}{2}) . Γ (\frac{1}{2})}{Γ (\frac{5}{2} + \frac{1}{2})}$ $= \frac{625}{2} \times \frac{\frac{3 \sqrt{π}}{4} \times \sqrt{π}}{Γ (3)} = \frac{625 π}{8} \times \frac{3}{2} (as$ $Γ (3) = 2!)$ $Thus, F = \frac{1875 π}{16}$
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