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IntegrationQuestion and Answers: Page 222

Question Number 68532 Answers: 0 Comments: 0

Question Number 68481 Answers: 0 Comments: 6

I=∫_0 ^( 1) (√((c−x^2 )/(x(1−x^2 ))))dx (c >1)

$${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{{c}−{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}{dx}\:\:\:\:\:\:\left({c}\:>\mathrm{1}\right) \\ $$

Question Number 68470 Answers: 0 Comments: 3

find the value of ∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$

Question Number 68434 Answers: 1 Comments: 0

Question Number 68409 Answers: 1 Comments: 3

calculate ∫_0 ^(+∞) ((arctan(x^2 ))/(1+x^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 68370 Answers: 1 Comments: 0

Question Number 68316 Answers: 0 Comments: 1

∫(4sin 3x+(e^(4x) /4))

$$\int\left(\mathrm{4sin}\:\mathrm{3}{x}+\frac{{e}^{\mathrm{4}{x}} }{\mathrm{4}}\right) \\ $$

Question Number 68313 Answers: 0 Comments: 1

∫(1−(6/x)+(2/x^2 )+(√x))

$$\int\left(\mathrm{1}−\frac{\mathrm{6}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }+\sqrt{{x}}\right) \\ $$

Question Number 68271 Answers: 0 Comments: 0

Find J=∫_0 ^1 ((W(−ulnu))/(ulnu)) du when W is the lambert function

$$\:\:{Find}\:\:{J}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{W}\left(−{ulnu}\right)}{{ulnu}}\:{du}\:\:\:\:{when}\:\:{W}\:{is}\:{the}\:{lambert}\:{function} \\ $$

Question Number 68241 Answers: 0 Comments: 1

calculate ∫∫_w (x^2 −2y^2 )(√(x^2 +3y^2 ))dxdy with w ={(x,y)∈R^2 / 0≤x≤1 and 1≤y≤2}

$${calculate}\:\int\int_{{w}} \:\:\:\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }{dxdy} \\ $$$${with}\:{w}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:{and}\:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\right\} \\ $$

Question Number 68220 Answers: 0 Comments: 0

Let consider (a_n )_n and (u_n )_n two reals sequence defined such as a_0 =1 , ∀ n>1 a_(n+1) =Σ_(p=0) ^n a_p a_(n−p) and Σ_(p=0) ^n a_p u_(n−p) =0 Part1 1)Express ∀ n >1 a_n in terms of n 2) Find the largest domain of convergence of the integer serie {a_n x^n } 3)Determinate ∀ x∈D the sum f(x) of {a_n x^n } 4)Find the radius of convergence of the serie {u_n x^n } 5) Give the relation that between the sum S(x) of the second serie and (x/(f(x))) 6) Can you developp in integer serie g(x)=((πx)/(tan(πx))) Part2 Now do the part 1 but in the order 2)−1)−3)−4)−5)−6)

$$\:\:\:{Let}\:{consider}\:\left({a}_{{n}} \right)_{{n}} \:{and}\:\left({u}_{{n}} \right)_{{n}} \:{two}\:{reals}\:\:{sequence}\:\: \\ $$$${defined}\:{such}\:{as}\:\:\:{a}_{\mathrm{0}} =\mathrm{1}\:,\:\forall\:{n}>\mathrm{1}\:\:{a}_{{n}+\mathrm{1}} =\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{p}} {a}_{{n}−{p}} \:\:\:{and}\:\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{p}} {u}_{{n}−{p}} =\mathrm{0} \\ $$$${Part}\mathrm{1} \\ $$$$\left.\mathrm{1}\right){Express}\:\:\forall\:{n}\:>\mathrm{1}\:\:\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{Find}\:{the}\:{largest}\:{domain}\:{of}\:{convergence}\:{of}\:{the}\:{integer}\:{serie}\:\left\{{a}_{{n}} {x}^{{n}} \right\} \\ $$$$\left.\mathrm{3}\right){Determinate}\:\forall\:{x}\in{D}\:{the}\:{sum}\:{f}\left({x}\right)\:{of}\:\left\{{a}_{{n}} {x}^{{n}} \right\} \\ $$$$\left.\mathrm{4}\right){Find}\:{the}\:{radius}\:{of}\:{convergence}\:{of}\:{the}\:{serie}\:\left\{{u}_{{n}} {x}^{{n}} \right\}\: \\ $$$$\left.\mathrm{5}\right)\:{Give}\:{the}\:{relation}\:{that}\:{between}\:{the}\:{sum}\:{S}\left({x}\right)\:{of}\:{the}\:{second}\:{serie}\:{and}\:\frac{{x}}{{f}\left({x}\right)}\: \\ $$$$\left.\mathrm{6}\right)\:{Can}\:{you}\:{developp}\:{in}\:{integer}\:{serie}\:\:{g}\left({x}\right)=\frac{\pi{x}}{{tan}\left(\pi{x}\right)} \\ $$$${Part}\mathrm{2} \\ $$$$\left.{N}\left.{o}\left.{w}\left.\:\left.{d}\left.{o}\:\:{the}\:{part}\:\mathrm{1}\:\:\:{but}\:{in}\:{the}\:{order}\:\:\mathrm{2}\right)−\mathrm{1}\right)−\mathrm{3}\right)−\mathrm{4}\right)−\mathrm{5}\right)−\mathrm{6}\right) \\ $$

Question Number 68219 Answers: 1 Comments: 0

Question Number 68149 Answers: 0 Comments: 2

Explicit f(a)=Σ_(n=1) ^∞ (((−1)^n )/(n(an+1)))

$$\:{Explicit}\:\:\:{f}\left({a}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({an}+\mathrm{1}\right)}\:\:\:\: \\ $$

Question Number 68145 Answers: 0 Comments: 5

Find the arc length, given the curve x(t) = sin (πt), y(t) = t , 0 ≤ t ≤ 1

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{length},\:\mathrm{given}\:\mathrm{the}\:\mathrm{curve} \\ $$$${x}\left({t}\right)\:=\:\mathrm{sin}\:\left(\pi{t}\right),\:\:{y}\left({t}\right)\:=\:{t}\:,\:\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{1} \\ $$

Question Number 68141 Answers: 0 Comments: 1

the 2 formulas for solving ∫(dx/(x^3 +px+q)) with “nasty” solutions of x^3 +px+q=0 with p, q ∈R case 1 D=(p^3 /(27))+(q^2 /4)>0 ⇒ x^3 +px+q=0 has got 1 real and 2 conjugated complex solutions u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) ∧v=((−(q/2)−p(√((p^3 /(27))+(q^2 /4)))))^(1/3) x_1 =u+v x_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v x_3 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v α=u+v∧β=((√3)/2)(u−v) ⇔ u=(α/2)+(β/(√3))∧v=(α/2)−(β/(√3)) x_1 =α x_2 =−(α/2)+βi x_3 =−(α/2)−βi ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x^2 +αx+((α^2 +4β^2 )/4))))= =(1/(9α^2 +4β^2 ))(∫(dx/((x−α)))−∫((x+2α)/(x^2 +αx+((α^2 +4β^2 )/4)))dx)= =(1/(9α^2 +4β^2 ))(ln ∣x−α∣ −(1/2)ln ((2x+α)^2 +4β^2 ) −((3α)/(2β))arctan ((2x+α)/(2β))) +C ...now calculate the constants case 2 D=(p^3 /(27))+(q^2 /4)<0 ⇒ x^3 +px+q=0 has got 3 real solutions x_k =(2/3)(√(−3p)) sin (((2π)/3)k+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 ))))) with k=0, 1, 2 let x_1 =α, x_2 =β, x_3 =γ ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x−β)(x−γ)))= =(1/((α−β)(α−γ)))∫(dx/(x−α))+(1/((β−α)(β−γ)))∫(dx/(x−β))+(1/((γ−α)(γ−β)))∫(dx/(x−γ))= =((ln ∣x−α∣)/((α−β)(α−γ)))+((ln ∣x−β∣)/((β−α)(β−γ)))+((ln ∣x−γ∣)/((γ−α)(γ−β)))+C ...now calculate the constants

$$\mathrm{the}\:\mathrm{2}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{solving}\:\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}\:\mathrm{with} \\ $$$$``\mathrm{nasty}''\:\mathrm{solutions}\:\mathrm{of}\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{with}\:{p},\:{q}\:\in\mathbb{R} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}\wedge{v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−{p}\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$${x}_{\mathrm{1}} ={u}+{v} \\ $$$${x}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${x}_{\mathrm{3}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\alpha={u}+{v}\wedge\beta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({u}−{v}\right)\:\Leftrightarrow\:{u}=\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\sqrt{\mathrm{3}}}\wedge{v}=\frac{\alpha}{\mathrm{2}}−\frac{\beta}{\sqrt{\mathrm{3}}} \\ $$$${x}_{\mathrm{1}} =\alpha \\ $$$${x}_{\mathrm{2}} =−\frac{\alpha}{\mathrm{2}}+\beta\mathrm{i} \\ $$$${x}_{\mathrm{3}} =−\frac{\alpha}{\mathrm{2}}−\beta\mathrm{i} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\int\frac{{dx}}{\left({x}−\alpha\right)}−\int\frac{{x}+\mathrm{2}\alpha}{{x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}}{dx}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\mathrm{ln}\:\mid{x}−\alpha\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\left(\mathrm{2}{x}+\alpha\right)^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} \right)\:−\frac{\mathrm{3}\alpha}{\mathrm{2}\beta}\mathrm{arctan}\:\frac{\mathrm{2}{x}+\alpha}{\mathrm{2}\beta}\right)\:+{C} \\ $$$$...\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions} \\ $$$${x}_{{k}} =\frac{\mathrm{2}}{\mathrm{3}}\sqrt{−\mathrm{3}{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{k}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}\right)\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{let}\:{x}_{\mathrm{1}} =\alpha,\:{x}_{\mathrm{2}} =\beta,\:{x}_{\mathrm{3}} =\gamma \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)}= \\ $$$$=\frac{\mathrm{1}}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}\int\frac{{dx}}{{x}−\alpha}+\frac{\mathrm{1}}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}\int\frac{{dx}}{{x}−\beta}+\frac{\mathrm{1}}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}\int\frac{{dx}}{{x}−\gamma}= \\ $$$$=\frac{\mathrm{ln}\:\mid{x}−\alpha\mid}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\beta\mid}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\gamma\mid}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}+{C} \\ $$$$...\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$

Question Number 68116 Answers: 1 Comments: 1

∫(dx/(sin2x−sec(x)))

$$\int\frac{{dx}}{{sin}\mathrm{2}{x}−{sec}\left({x}\right)} \\ $$

Question Number 68100 Answers: 0 Comments: 4

Find K=∫_0 ^1 ((ln(1−t+t^2 ))/t) dt

$${Find}\:\:{K}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)}{{t}}\:{dt}\:\:\:\:\: \\ $$

Question Number 68095 Answers: 1 Comments: 0

∫(√x)/1+((x ))^(1/3) dx

$$\int\sqrt{{x}}/\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}\:}\:{dx} \\ $$

Question Number 68094 Answers: 1 Comments: 0

∫dx/(((π+e)^x^2 ))^(1/x)