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Question Number 221521 Answers: 0 Comments: 1

Question Number 221513 Answers: 0 Comments: 0

Question Number 221541 Answers: 1 Comments: 0

∫_0 ^( π) tan(4!)^e^(3x^2 + 2x) .dx

$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{tan}\left(\mathrm{4}!\right)^{{e}^{\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{2}{x}} } .{dx} \\ $$

Question Number 221504 Answers: 1 Comments: 0

Find x if (x^4 /4^x )=(4^x /x^4 ) . x∈R

$${Find}\:{x}\:{if}\:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{{x}^{\mathrm{4}} }\:\:.\:\:\:{x}\in\mathbb{R} \\ $$

Question Number 221501 Answers: 3 Comments: 0

solve for x: (√(a−(√(a+x))))+(√(a+(√(a−x))))=2x it′s possible to solve for a but x seems impossible to me

$${solve}\:{for}\:\mathrm{x}: \\ $$$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}=\mathrm{2x} \\ $$$${it}'{s}\:{possible}\:{to}\:{solve}\:{for}\:\mathrm{a}\:{but}\:\mathrm{x}\:{seems} \\ $$$${impossible}\:{to}\:{me} \\ $$

Question Number 221500 Answers: 0 Comments: 0

Ip = ((p1q0)/(p0q0))×100

$$\boldsymbol{{Ip}}\:=\:\frac{\mathrm{p1q0}}{\mathrm{p0q0}}×\mathrm{100} \\ $$

Question Number 221498 Answers: 1 Comments: 0

Solve differantial Equation ((d )/dt)[((dy(t))/dt)]+ty(t)=0

$$\mathrm{Solve}\:\mathrm{differantial}\:\mathrm{Equation} \\ $$$$\frac{{d}\:\:}{{dt}}\left[\frac{{dy}\left({t}\right)}{{dt}}\right]+{ty}\left({t}\right)=\mathrm{0} \\ $$

Question Number 221495 Answers: 0 Comments: 1

Question Number 221484 Answers: 1 Comments: 0

I = ∫_( [0,1]^( 4) ) (1/( (√(1−x^2 y^2 −x^2 z^2 −x^2 w^2 −y^2 z^2 −y^2 w^2 −z^2 w^2 )))) dxdydzdw

$$ \\ $$$$\:\:\:{I}\:=\:\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\:\mathrm{4}} } \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{2}} {z}^{\mathrm{2}} −{x}^{\mathrm{2}} {w}^{\mathrm{2}} −{y}^{\mathrm{2}} {z}^{\mathrm{2}} −{y}^{\mathrm{2}} {w}^{\mathrm{2}} −{z}^{\mathrm{2}} {w}^{\mathrm{2}} }}\:\mathrm{d}{x}\mathrm{d}{y}\mathrm{d}{z}\mathrm{d}{w}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221483 Answers: 0 Comments: 1

I(α) = ∫∫∫_( [0,1]^3 ) ((ln(1 + α(xy + yz + zx)))/(1 − xyz )) dxdydz for α ∈ (0,1)

$$ \\ $$$$\:\:\:\:{I}\left(\alpha\right)\:=\:\int\int\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\alpha\left({xy}\:+\:{yz}\:+\:{zx}\right)\right)}{\mathrm{1}\:−\:{xyz}\:}\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\alpha\:\in\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 221481 Answers: 0 Comments: 0

prove:Σ_(k=0) ^n ((k!(n−k)!)/(n!))=−((i2^(n−1) Γ(n+2)(π−iB_2 (n+2,0)))/(n!))

$$\mathrm{prove}:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{k}!\left({n}−{k}\right)!}{{n}!}=−\frac{{i}\mathrm{2}^{{n}−\mathrm{1}} \Gamma\left({n}+\mathrm{2}\right)\left(\pi−{iB}_{\mathrm{2}} \left({n}+\mathrm{2},\mathrm{0}\right)\right)}{{n}!} \\ $$

Question Number 221469 Answers: 1 Comments: 0

Prove; ∫_0 ^( 1) ((((√(x )) ln (√x))^2 )/((1−x^2 )^2 + 4x^2 + 2x^4 + 3x^6 + 2x^8 + x^(10) )) dx = ((21)/(1024)) ζ(3) − (π^3 /(1024)) + (π^3 /(324(√3)))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\left(\sqrt{{x}\:}\:\mathrm{ln}\:\sqrt{{x}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{4}} \:+\:\mathrm{3}{x}^{\mathrm{6}} \:+\:\mathrm{2}{x}^{\mathrm{8}} \:+\:{x}^{\mathrm{10}} }\:\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{21}}{\mathrm{1024}}\:\zeta\left(\mathrm{3}\right)\:−\:\frac{\pi^{\mathrm{3}} }{\mathrm{1024}}\:+\:\frac{\pi^{\mathrm{3}} }{\mathrm{324}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Question Number 221454 Answers: 2 Comments: 0

Prove ∫^( +∞) _( 1) (x/((−1 + 3x^2 − 3x^4 + 2x^6 )(ln(x−1) − 2 ln x + ln(x+1)))) dx = − ((ln 3)/4)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove} \\ $$$$\:\:\underset{\:\mathrm{1}} {\int}^{\:+\infty} \:\frac{{x}}{\left(−\mathrm{1}\:+\:\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{6}} \right)\left(\mathrm{ln}\left({x}−\mathrm{1}\right)\:−\:\mathrm{2}\:\mathrm{ln}\:{x}\:+\:\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}\:\:\mathrm{d}{x}\:=\:−\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221453 Answers: 2 Comments: 0

Question Number 221447 Answers: 1 Comments: 0

if Π_(i=1) ^n (x + r_i ) ≡ Σ_(j=0) ^n a_j x^(n−i) show that ; Σ_(i=1) ^n tan^(−1) r_i = tan^(−1) ((a_1 + a_3 + a_5 − ∙∙∙)/(a_0 − a_2 + a_4 −∙∙∙)) and Σ_(i=1) ^n tanh^(−1) r_i = tanh^(−1) ((a_1 + a_3 + a_5 + ∙∙∙)/(a_0 + a_2 + a_4 + ∙∙∙))

$$ \\ $$$$\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:\left({x}\:+\:{r}_{{i}} \right)\:\equiv\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\:{a}_{{j}} {x}^{{n}−{i}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:; \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} +\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:−\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:−\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:−\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and} \\ $$$$\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tanh}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tanh}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:+\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:+\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:+\:\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221444 Answers: 0 Comments: 3

Question Number 221438 Answers: 0 Comments: 0

∫_( 1) ^∞ ((ln(ln x))/(1−2x cos θ + x^2 )) dx ; for all θ ∈ (−π , π)

$$ \\ $$$$\:\:\:\int_{\:\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta\:+\:{x}^{\mathrm{2}} }\:{dx}\:;\:\mathrm{for}\:\mathrm{all}\:\theta\:\in\:\left(−\pi\:,\:\pi\right)\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221430 Answers: 1 Comments: 0

Question Number 221417 Answers: 0 Comments: 0

Let S be delimited by the equations x=0; y=0 ; z=0 and x+y+z=0 Find the flux of vector field V(x,y,z)=(x,y,x^2 +y^2 ) through S

$${Let}\:\:{S}\:{be}\:{delimited}\:{by}\:{the}\:{equations}\: \\ $$$${x}=\mathrm{0};\:{y}=\mathrm{0}\:;\:{z}=\mathrm{0}\:{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${Find}\:{the}\:{flux}\:{of}\:{vector}\:{field}\: \\ $$$${V}\left({x},{y},{z}\right)=\left({x},{y},{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:{through}\:{S} \\ $$

Question Number 221416 Answers: 0 Comments: 0

ex3. prove f^((n)) (α)=((n!)/(2πi)) ∮_( ∂S) ((f(z))/((z−α)^(n+1) )) dz ex4. Let z_0 be any point interior to a positively oriented simple closed contour C show that a. ∮_C (dz/(z−z_0 ))=2πi b. ∮_( C) (dz/((z−z_0 )^(n+1) ))=0 , n∈R^+ ex 5. Let C be any simple closed contour, described in the positive sense in the z plane and write g(z)= ∮_( C) ((s^3 +2s)/((s−z)^3 )) ds show that g(z)=6πi when z is inside C and that g(z)=0 when z is outside

$$\mathrm{ex3}. \\ $$$$\mathrm{prove} \\ $$$${f}^{\left({n}\right)} \left(\alpha\right)=\frac{{n}!}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\partial{S}} \:\frac{{f}\left({z}\right)}{\left({z}−\alpha\right)^{{n}+\mathrm{1}} }\:\mathrm{d}{z} \\ $$$$\mathrm{ex4}. \\ $$$$\mathrm{Let}\:{z}_{\mathrm{0}} \:\mathrm{be}\:\mathrm{any}\:\mathrm{point}\:\mathrm{interior}\:\mathrm{to}\:\mathrm{a}\:\mathrm{positively} \\ $$$$\mathrm{oriented}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}\:\mathcal{C} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\oint_{{C}} \:\frac{\mathrm{d}{z}}{{z}−{z}_{\mathrm{0}} }=\mathrm{2}\pi\boldsymbol{{i}} \\ $$$$\mathrm{b}.\:\oint_{\:{C}} \frac{\mathrm{d}{z}}{\left({z}−{z}_{\mathrm{0}} \right)^{{n}+\mathrm{1}} }=\mathrm{0}\:,\:{n}\in\mathbb{R}^{+} \\ $$$$\mathrm{ex}\:\mathrm{5}. \\ $$$$\mathrm{Let}\:\mathcal{C}\:\mathrm{be}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}, \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{sense}\:\mathrm{in}\:\mathrm{the}\:{z}\:\mathrm{plane} \\ $$$$\mathrm{and}\:\mathrm{write}\: \\ $$$$\mathrm{g}\left({z}\right)=\:\oint_{\:\mathcal{C}} \:\frac{{s}^{\mathrm{3}} +\mathrm{2}{s}}{\left({s}−{z}\right)^{\mathrm{3}} }\:\mathrm{d}{s} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{6}\pi\boldsymbol{{i}}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{inside}\:\mathcal{C}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{0}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{outside} \\ $$

Question Number 221415 Answers: 1 Comments: 0

∫ dz [−(1/π)((2/z))^ν ∙Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) +(2/π)ln((1/2)z)J_ν (z)−(1/π)((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (ψ^((0)) (k+ν+1)+ψ^((0)) (k+1)))/(k!(k+ν)!))((z/2))^(2k) ]

$$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$

Question Number 221413 Answers: 7 Comments: 0

Question Number 221411 Answers: 1 Comments: 0

Question Number 221407 Answers: 0 Comments: 0

A and B are two angles such that 0^0 <B<A<90^0 then prove geometrycaly that cos (A+B)=cos Acos B−sin Asin B

$${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$

Question Number 221406 Answers: 1 Comments: 1

Question Number 221405 Answers: 0 Comments: 0

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