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Question Number 68129 Answers: 0 Comments: 0

prove that πcotan(απ)=(1/α) +Σ_(n=1) ^∞ ((2α)/(α^2 −n^2 )) with α ∈R−Z . prove also that for t≠0 cotan(t) =(1/t) +Σ_(n=1) ^∞ ((2t)/(t^2 −n^2 π^2 ))

$${prove}\:{that}\:\pi{cotan}\left(\alpha\pi\right)=\frac{\mathrm{1}}{\alpha}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$${with}\:\alpha\:\in{R}−{Z}\:\:. \\ $$$${prove}\:{also}\:{that}\:\:\:{for}\:{t}\neq\mathrm{0} \\ $$$${cotan}\left({t}\right)\:=\frac{\mathrm{1}}{{t}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −{n}^{\mathrm{2}} \pi^{\mathrm{2}} } \\ $$

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 68122 Answers: 1 Comments: 1

Question Number 68116 Answers: 1 Comments: 1

∫(dx/(sin2x−sec(x)))

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

Question Number 68113 Answers: 1 Comments: 0

Solve y.y′′ = 3(y′)^2

$$\mathrm{Solve} \\ $$$${y}.{y}''\:=\:\mathrm{3}\left({y}'\right)^{\mathrm{2}} \\ $$

Question Number 68110 Answers: 0 Comments: 2

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 68096 Answers: 0 Comments: 0

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)

$$\int{dx}/\sqrt[{{x}}]{\left(\pi+{e}\right)^{{x}^{\mathrm{2}} } }\: \\ $$

Question Number 68092 Answers: 0 Comments: 4

Question Number 68087 Answers: 0 Comments: 2

Hello friends! I ask name of a high level book of integral calculus.

$${Hello}\:{friends}!\:{I}\:{ask}\:{name}\:{of}\:{a}\:{high} \\ $$$${level}\:{book}\:{of}\:{integral}\:{calculus}. \\ $$

Question Number 68079 Answers: 1 Comments: 0

Question Number 68078 Answers: 1 Comments: 0

Question Number 68077 Answers: 0 Comments: 0

Question Number 68073 Answers: 0 Comments: 1

A straight rod AB which is 60cm long,is in equilibrum when horizontal and supported at a point C,10cm from A, with masses 6kg and 1kg attached to the rod at A and B respectively.It is also in equilibrum and horizontal when supported at another pivott at its mid- point,with masses of 2kg and 5kg attatched at A and B respectively.Find the mass of the rod amd its C.G from point A.

$$\:{A}\:{straight}\:{rod}\:{AB}\:{which}\:{is}\:\mathrm{60}{cm}\:{long},{is} \\ $$$${in}\:{equilibrum}\:{when}\:{horizontal}\:{and} \\ $$$${supported}\:{at}\:{a}\:{point}\:{C},\mathrm{10}{cm}\:{from}\:{A}, \\ $$$${with}\:{masses}\:\mathrm{6}{kg}\:{and}\:\mathrm{1}{kg}\:{attached}\:{to}\:{the} \\ $$$${rod}\:{at}\:{A}\:{and}\:{B}\:{respectively}.{It}\:{is}\:{also}\:{in} \\ $$$${equilibrum}\:{and}\:{horizontal}\:{when}\: \\ $$$${supported}\:{at}\:{another}\:{pivott}\:{at}\:{its}\:{mid}- \\ $$$${point},{with}\:{masses}\:{of}\:\mathrm{2}{kg}\:{and}\:\mathrm{5}{kg}\: \\ $$$${attatched}\:{at}\:{A}\:{and}\:{B}\:{respectively}.{Find} \\ $$$${the}\:{mass}\:{of}\:{the}\:{rod}\:{amd}\:{its}\:{C}.{G}\:{from} \\ $$$${point}\:{A}. \\ $$

Question Number 68068 Answers: 0 Comments: 2

find e^(1/ln2) =?

$${find}\:{e}^{\mathrm{1}/{ln}\mathrm{2}} \:\:=? \\ $$

Question Number 68062 Answers: 0 Comments: 1