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Question Number 141098    Answers: 0   Comments: 0

Question Number 141087    Answers: 2   Comments: 0

hi, masters ! look at this thing carefully : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ................. find the line and the column where the number 795471 will appear !

$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{masters}}\:! \\ $$$$\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{thing}}\:\boldsymbol{\mathrm{carefully}}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}\:\mathrm{8}\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}\:\mathrm{11}\:\mathrm{12}\:\mathrm{13}\:\mathrm{14}\:\mathrm{15}\:\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{17}\:\mathrm{18}\:\mathrm{19}\:\mathrm{20}\:................. \\ $$$$\:\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{line}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{column}}\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{number}}\:\mathrm{795471}\:\boldsymbol{\mathrm{will}}\:\boldsymbol{\mathrm{appear}}\:! \\ $$

Question Number 141085    Answers: 2   Comments: 0

Question Number 141062    Answers: 1   Comments: 4

prove that:: φ:=∫_0 ^( ∞) ((e^(cos(x)) sin(sin(x)))/x) dx=(π/2)(e−1)

$$\:\:\:\:\: \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{cos}\left({x}\right)} {sin}\left({sin}\left({x}\right)\right)}{{x}}\:{dx}=\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$

Question Number 141057    Answers: 2   Comments: 0

.....Nice ...... ......Calculus..... prove that: Ω(x):=Σ_(n=1) ^∞ a^n .((sin(nx))/(n!))=e^(acos(x)) sin(asin(x)) ....m.n

$$ \\ $$$$\:\:\:\:\:\:\:.....\mathscr{N}{ice}\:......\:\:......\mathscr{C}{alculus}..... \\ $$$$\:\:\:{prove}\:{that}: \\ $$$$\:\Omega\left({x}\right):=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}^{{n}} .\frac{{sin}\left({nx}\right)}{{n}!}={e}^{{acos}\left({x}\right)} {sin}\left({asin}\left({x}\right)\right) \\ $$$$\:\:\:....{m}.{n} \\ $$

Question Number 141076    Answers: 1   Comments: 0

t+x=(c/2) t^2 +x^4 =(c^2 /4) find x or t . Given 0<c<(2/(3(√3)))

$${t}+{x}=\frac{{c}}{\mathrm{2}} \\ $$$$\:{t}^{\mathrm{2}} +{x}^{\mathrm{4}} =\frac{{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${find}\:{x}\:{or}\:{t}\:.\:{Given}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\: \\ $$

Question Number 141050    Answers: 1   Comments: 0

Question Number 141077    Answers: 2   Comments: 0

Calculate the lim_(x→1) ((((√x^x ) − (√x))/(x^x − x))).

$$\:{Calculate}\:\:{the}\:\:\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\sqrt{{x}^{{x}} }\:−\:\sqrt{{x}}}{{x}^{{x}} \:−\:{x}}\right). \\ $$

Question Number 141082    Answers: 3   Comments: 0

........ nice ....... calculus ........ 𝛗:=Σ_(n=2) ^∞ ((ζ ( n ))/(n . 4^n ))=?

$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:........\:{nice}\:\:.......\:\:{calculus}\:........ \\ $$$$\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\zeta\:\left(\:{n}\:\right)}{{n}\:.\:\mathrm{4}^{{n}} }=? \\ $$$$ \\ $$

Question Number 141045    Answers: 0   Comments: 4

Question Number 141034    Answers: 0   Comments: 2

I don′t recover my old phone documents. please advise me in briefly how to restore my old phone documents in my new phone. plese help me.

$${I}\:{don}'{t}\:{recover}\:{my}\:{old}\:{phone}\:{documents}. \\ $$$${please}\:{advise}\:{me}\:{in}\:{briefly}\:{how}\:{to}\:{restore} \\ $$$$\:{my}\:{old}\:{phone}\:{documents}\:{in}\:{my}\:{new} \\ $$$$\:{phone}. \\ $$$$ \\ $$$${plese}\:{help}\:{me}. \\ $$

Question Number 141031    Answers: 0   Comments: 0

∫_(π/6) ^( π/3) (√(1+((cos^2 x)/(sin x)))) dx ?

$$\:\:\:\:\:\:\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{3}} \:\sqrt{\mathrm{1}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}}\:{dx}\:?\: \\ $$

Question Number 141030    Answers: 1   Comments: 1

→⟨Σ_(n=1) ^∞ (1/(9n^2 +3n)) =? ⟩←

$$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$

Question Number 141024    Answers: 3   Comments: 0

Given X=((√(√x))+(1/( (√(√x)))))^n . What is the coefficient of x^(5−(n/4) ?) propositions: a. 5((n!)/(10!)) b. ((n!)/(5!)) c. ((n),((10)) ) d. ((n),(5) )

$${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}−\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$

Question Number 141017    Answers: 2   Comments: 1

Question Number 141004    Answers: 0   Comments: 0

Let 0 ≤ a,b < 1. Prove that (1/4)∙(((2−a)(2−b))/((1−a)(1−b))) ≥ ((4+a+b)/(4−a−b))

$$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140998    Answers: 1   Comments: 0

Question Number 140997    Answers: 0   Comments: 1

Let a,b,c ≥ 0. Prove that (1/8)∙(((2+a)(2+b)(2+c))/((1+a)(1+b)(1+c))) ≥ ((4−a−b−c)/(4+a+b+c))

$$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140996    Answers: 1   Comments: 0

Evaluation of :: Ω :=Σ_(n=0) ^∞ (((−1)^n )/(1+n^2 )) solution:: Ω:=1+Σ_(n=1) ^∞ (((−1)^n )/(n^2 −i^2 )) =(1/(2i)){Σ_(n=1) ^∞ (((−1)^n )/(n−i))−(((−1)^n )/(n+i))} :=1+(1/(2i)) (Φ−Ψ) where Φ:=Σ_(n=1) ^∞ (((−1)^n )/(n−i)) and Ψ :=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) Φ:=Σ_(n=1) ^∞ (1/(2n−k)) −Σ_(n=1) ^∞ (1/(2n−1−i)) :=(1/2){Σ_(n=1) ^∞ (1/(n−(i/2)))−Σ_(n=1) ^∞ (1/(n−((1+i)/2)))} :=(1/2){ψ(1−((1+i)/2))−ψ(1−(i/2))} :=(1/2)(ψ(((1−i)/2))−ψ(1−(i/2))).... Ψ:=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) =Σ_(n=1) ^∞ (1/(2n+i))−Σ_(n=1) ^∞ (1/(2n−1+i)) :=(1/2){Σ_(n=1) ^∞ (1/(n+(i/2)))−Σ_(n=1) ^∞ (1/(n+((i−1)/2)))} :=(1/2)(ψ(((1+i)/2))−ψ(1+(i/2))) .... Φ−Ψ:=(1/2){ψ(((1−i)/2))−ψ(((1+i)/2))} +(1/2){ψ(1+(i/2))−ψ(1−(i/2))} :=(1/2)(−πcotπ(((1−i)/2)))+(1/2)((2/i)−πcot(π(i/2))) :=−(π/2)tan(((πi)/2))−(π/2)cot(((πi)/2))−i :=−i−(π/(sin(πi)))=−i−((2iπ)/(e^(−π) −e^π )) :=−i+πicsch(π) .... Ω :=1+(1/(2i))(−i+πicsch(π))=(1/2)+(π/2) csch(π) ... Ω:=(1/2)+(π/2) csch(π)....✓✓✓

$$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}}−\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\right\} \\ $$$$\:\:\:\:\:\:\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left(\Phi−\Psi\right)\:\:\:\:{where}\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}} \\ $$$$\:\:\:\:\:\:\:{and}\:\:\:\:\Psi\::=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−{k}}\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}−{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}+{i}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}−\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right).... \\ $$$$\:\:\:\:\Psi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{i}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}+{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}−\mathrm{1}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right)\:.... \\ $$$$\:\:\:\:\:\Phi−\Psi:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(−\pi{cot}\pi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{i}}−\pi{cot}\left(\pi\frac{{i}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=−\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi{i}}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi{i}}{\mathrm{2}}\right)−{i} \\ $$$$\:\:\:\::=−{i}−\frac{\pi}{{sin}\left(\pi{i}\right)}=−{i}−\frac{\mathrm{2}{i}\pi}{{e}^{−\pi} −{e}^{\pi} } \\ $$$$\:\:\:\::=−{i}+\pi{icsch}\left(\pi\right)\:.... \\ $$$$\:\:\:\:\:\:\Omega\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−{i}+\pi{icsch}\left(\pi\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)....\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 140988    Answers: 1   Comments: 0

∫((−csc^2 x)/((cscx+cotx)^3 ))dx

$$\int\frac{−{csc}^{\mathrm{2}} {x}}{\left({cscx}+{cotx}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 141012    Answers: 0   Comments: 0

s^2 (s+1)^2 +(c/2)s(s+1)^2 −(c^2 /2)(s+1) +(c^3 /2) = 0 solve for s in terms of 0<c<(2/(3(√3))) .

$$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

Question Number 140982    Answers: 0   Comments: 3

convergence and value of Σ_(n=1) ^∞ (n^n /((n!)^2 ))

$${convergence}\:{and}\:{value}\:{of} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{{n}} }{\left({n}!\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Question Number 140981    Answers: 1   Comments: 0

Let α≠1 and α^(13) =1. If a=α+α^3 +α^4 +α^(−4) +α^(−3) + α^(−1) and b=α^2 +α^5 +α^6 +α^(−6) +α^(−5) +α^(−2) then the quadratic equation whose roots are a and b is (A) x^2 +x+3=0 (B) x^2 +x+4=0 (C) x^2 +x−3=0 (D) x^2 +x−4=0

$$\mathrm{Let}\:\alpha\neq\mathrm{1}\:\mathrm{and}\:\alpha^{\mathrm{13}} =\mathrm{1}.\:\mathrm{If}\:{a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{−\mathrm{4}} +\alpha^{−\mathrm{3}} + \\ $$$$\alpha^{−\mathrm{1}} \:\mathrm{and}\:{b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{−\mathrm{6}} +\alpha^{−\mathrm{5}} +\alpha^{−\mathrm{2}} \:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{4}=\mathrm{0} \\ $$

Question Number 140978    Answers: 1   Comments: 0

calculate Σ_(n=0) ^∞ (1/((n!)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$

Question Number 140977    Answers: 2   Comments: 0

find ∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx with t≥0

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$

Question Number 140976    Answers: 1   Comments: 0

find ∫_0 ^π (dx/((2−cosx−sinx)^2 ))

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{2}−{cosx}−{sinx}\right)^{\mathrm{2}} } \\ $$$$ \\ $$

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