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Question Number 223055    Answers: 1   Comments: 2

Question Number 223054    Answers: 0   Comments: 0

find y y^dy = x^dx

$$\mathrm{find}\:{y} \\ $$$${y}^{{dy}} =\:{x}^{{dx}} \\ $$

Question Number 223090    Answers: 1   Comments: 1

∫_0 ^∞ (x^2 /((cosh(x^2 ))^2 ))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{cosh}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{2}} }\mathrm{d}{x} \\ $$

Question Number 223044    Answers: 0   Comments: 1

Question Number 223042    Answers: 1   Comments: 12

Question Number 223040    Answers: 0   Comments: 1

Question Number 223032    Answers: 1   Comments: 0

Π_(t=0) ^2^(2024) (4sin^2 ((tπ)/2^(2025) )−3)

$$ \\ $$$$\:\:\:\:\underset{{t}=\mathrm{0}} {\overset{\mathrm{2}^{\mathrm{2024}} } {\prod}}\left(\mathrm{4sin}^{\mathrm{2}} \frac{{t}\pi}{\mathrm{2}^{\mathrm{2025}} }−\mathrm{3}\right) \\ $$

Question Number 223025    Answers: 0   Comments: 0

Question Number 223007    Answers: 0   Comments: 6

everyone or Mr. Gaster ! Please help me how to sove the integral Because the integral is very crazy or very Complicated Problem; ∫_0 ^( 1) ln(x−1) ln(x+1) ln (x^2 + 1) tan^(−1) (x) dx =???

$$ \\ $$$$\:\:\:\:\:\:\mathrm{everyone}\:\mathrm{or}\:\mathrm{Mr}.\:\mathrm{Gaster}\:! \\ $$$$\:\:\:\:\:\:\:\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{how}\:\mathrm{to}\:\mathrm{sove}\:\mathrm{the}\:\mathrm{integral}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Because}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{very}\:\mathrm{crazy}\:\mathrm{or}\:\mathrm{very}\:\mathrm{Complicated} \\ $$$$\:\:\:\:\:\:\mathrm{Problem}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{ln}\left({x}−\mathrm{1}\right)\:\mathrm{ln}\left({x}+\mathrm{1}\right)\:\mathrm{ln}\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\mathrm{d}{x}\:=??? \\ $$$$ \\ $$

Question Number 223006    Answers: 1   Comments: 0

L { tsin((√t) )}=?

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathscr{L}\:\:\left\{\:{tsin}\left(\sqrt{{t}}\:\right)\right\}=? \\ $$

Question Number 223000    Answers: 1   Comments: 1

Π_(n=1) ^∞ ((1/e)(((n+1)/n))^n )^((−1)^n ) =((e∙(√π)∙(2)^(1/6) )/A^6 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{1}}{{e}}\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{{n}} \right)^{\left(−\mathrm{1}\right)^{{n}} } =\frac{{e}\centerdot\sqrt{\pi}\centerdot\sqrt[{\mathrm{6}}]{\mathrm{2}}}{{A}^{\mathrm{6}} } \\ $$

Question Number 222996    Answers: 2   Comments: 0

∫_0 ^1 (2−x^2 )^(3/2) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} {dx} \\ $$

Question Number 222995    Answers: 1   Comments: 0

Question Number 222993    Answers: 0   Comments: 1

Question Number 222974    Answers: 1   Comments: 0

Question Number 223034    Answers: 0   Comments: 0

∫((x arctanxdx)/( (√((1+x^2 )(1+k′^2 x^2 )))))=(1/k^2 )[F((π/4),k)−(π/(2(√(2(1+k′2)))))]

$$ \\ $$$$\int\frac{{x}\:\mathrm{arctan}{xdx}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{k}'^{\mathrm{2}} {x}^{\mathrm{2}} \right)}}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left[{F}\left(\frac{\pi}{\mathrm{4}},{k}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+{k}'\mathrm{2}\right)}}\right] \\ $$

Question Number 222961    Answers: 1   Comments: 0

Question Number 222958    Answers: 2   Comments: 6

Question Number 222955    Answers: 1   Comments: 1

Question Number 222954    Answers: 3   Comments: 2

$$ \\ $$

Question Number 222943    Answers: 4   Comments: 0

11. If (√5) = 2.236 and (√(10)) = 3.162 then the value of ((15)/( (√(10))+(√(20))+(√(40))−(√5)−(√(80)))) is (a) 5.398 (b) 4.398 (c) 3.398 (d) 6.398 12. If x=(((√3)+1)/3) then x^3 +(1/(x^3 ))=? (a) ((28(√3) +15)/8) (b) ((28(√3)−15)/8) (c) ((27(√3)−35)/4) (d) ((27(√3)+35)/4) 13. Simplify ((x^4 ((x^3_ ((x^2 (√x)))^(1/3) ))^(1/4) ))^(1/5) (a) x^((23)/(24)) (b) x^((23)/6) (c) x^(5/6) (d) x^((119)/(120)) 14. ((3+2(√5))/(4−2(√5)))=p+q(√5) where p and q are rational numbers then values of p and q respectively are (a) −8 −(7/2) (b) 8 −(7/2) (c) 4 7 (d) −4 −7 16. 2.6^− − 0.8^− 2^− is equal to (a) ((181)/(999)) (b) ((82)/(99)) (c) ((182)/(99)) (d_(If) ) Non of these 17. If x=3(√5)+2(√2) and y=3(√5)−2(√2) then the value of (x^2 −y^2 )^2 is (a) 5760

$$\mathrm{11}.\:\mathrm{If}\:\sqrt{\mathrm{5}}\:=\:\mathrm{2}.\mathrm{236}\:\mathrm{and}\:\sqrt{\mathrm{10}}\:=\:\mathrm{3}.\mathrm{162}\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{15}}{\:\sqrt{\mathrm{10}}+\sqrt{\mathrm{20}}+\sqrt{\mathrm{40}}−\sqrt{\mathrm{5}}−\sqrt{\mathrm{80}}}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{5}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{4}.\mathrm{398}\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{3}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{6}.\mathrm{398} \\ $$$$\mathrm{12}.\:\mathrm{If}\:\mathrm{x}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}}\:\:\mathrm{then}\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \:}=? \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}\:+\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}−\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}−\mathrm{35}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}+\mathrm{35}}{\mathrm{4}} \\ $$$$\mathrm{13}.\:\mathrm{Simplify}\:\:\sqrt[{\mathrm{5}}]{\mathrm{x}^{\mathrm{4}} \sqrt[{\mathrm{4}}]{\mathrm{x}^{\mathrm{3}_{\:} } \sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}}} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{x}^{\frac{\mathrm{23}}{\mathrm{24}}} \:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{x}^{\frac{\mathrm{23}}{\mathrm{6}}} \:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{6}}} \:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{x}^{\frac{\mathrm{119}}{\mathrm{120}}} \\ $$$$\mathrm{14}.\: \\ $$$$ \\ $$$$ \: \\ $$$$ \\ $$$$ \\ $$$$ \:\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}}=\mathrm{p}+\mathrm{q}\sqrt{\mathrm{5}}\:\:\mathrm{where}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{rational}\:\mathrm{numbers}\:\mathrm{then}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{respectively}\:\mathrm{are} \\ $$$$\left(\mathrm{a}\right)\:−\mathrm{8}\:\:\:−\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{8}\:\:\:\:−\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:−\mathrm{4}\:\:\:\:\:−\mathrm{7} \\ $$$$\mathrm{16}.\:\mathrm{2}.\overset{−} {\mathrm{6}}\:−\:\mathrm{0}.\overset{−} {\mathrm{8}}\overset{−} {\mathrm{2}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{181}}{\mathrm{999}}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{82}}{\mathrm{99}}\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{182}}{\mathrm{99}}\:\:\:\:\:\:\:\:\:\:\left(\underset{\mathrm{If}} {\mathrm{d}}\right)\:\mathrm{Non}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{17}.\:\mathrm{If}\:\mathrm{x}=\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{y}=\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{5760}\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

Question Number 222937    Answers: 1   Comments: 0

Prove that the following identity holds : Σ_(λ∈Z[i]) ((1/((1 + 3λ)^3 ))) = ((π^(9/2) (√(1 + (√(3 )))))/(2^(5/4) 3^(27/8) Γ((3/4))^6 )) Where Z[i] = {a + bi : a,b ∈ Z} denotes gaussian integers !

$$ \\ $$$$\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{identity}\:\mathrm{holds}\::\:\:\:\: \\ $$$$\:\:\:\underset{\lambda\in\mathbb{Z}\left[{i}\right]} {\sum}\:\left(\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{3}\lambda\right)^{\mathrm{3}} }\right)\:=\:\frac{\pi^{\mathrm{9}/\mathrm{2}} \:\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{3}\:}}}{\mathrm{2}^{\mathrm{5}/\mathrm{4}} \:\:\mathrm{3}^{\mathrm{27}/\mathrm{8}} \:\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{6}} \:}\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{Where}\:\mathbb{Z}\left[{i}\right]\:=\:\left\{{a}\:+\:{bi}\::\:{a},{b}\:\in\:\mathbb{Z}\right\}\:\mathrm{denotes}\:\mathrm{gaussian}\:\mathrm{integers}\:!\:\:\:\:\:\:\:\:\: \\ $$$$\: \\ $$

Question Number 222929    Answers: 0   Comments: 0

Question Number 222927    Answers: 0   Comments: 0

∫∫∫_V ▽∙FdV=∫∫_∂V F∙dS F=−▽φ ▽∙(−▽φ)=−▽^2 φ ∫∫∫_V (−▽^2 φ)dV=∫∫_∂V (−▽φ)∙dS −∫∫∫_V ▽^2 φdV=−∫∫_∂V ▽φ∙dS ∫∫∫_V ▽^2 φdV=∫∫_∂V ▽φ∙dS ρ(r)=Σ_i q_i δ^((3)) −(r−r_i ) ▽^2 ((1/(∣r−r′∣)))=−4πδ^((3)) (r−r′) φ(r)=∫∫∫_V ((ρ(r′))/(4π∣r−r′∣))dV′ ▽^2 φ=▽^2 ((1/(4π))∫∫∫_V ((ρ(r′))/(∣r−r′∣))dV′) =(1/(4π))∫∫∫_V ρ(r′)▽^2 ((1/(∣r−r′∣)))dV′ =(1/(4π))∫∫∫_V ρ(r′)(−4πδ^((3)) (r−r′))dV′ =−∫∫∫_V ρ(r′)δ^((3)) (r−r′)dV′ =−ρ(r) ▽^2 φ=−ρ

$$\int\int\int_{{V}} \bigtriangledown\centerdot\boldsymbol{\mathrm{F}}{dV}=\int\int_{\partial{V}} \boldsymbol{\mathrm{F}}\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$\boldsymbol{\mathrm{F}}=−\bigtriangledown\phi \\ $$$$\bigtriangledown\centerdot\left(−\bigtriangledown\phi\right)=−\bigtriangledown^{\mathrm{2}} \phi \\ $$$$\int\int\int_{{V}} \left(−\bigtriangledown^{\mathrm{2}} \phi\right){dV}=\int\int_{\partial{V}} \left(−\bigtriangledown\phi\right)\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$−\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi\mathrm{d}{V}=−\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi{dV}=\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\rho\left(\boldsymbol{{r}}\right)=\underset{{i}} {\sum}{q}_{{i}} \delta^{\left(\mathrm{3}\right)} −\left(\boldsymbol{{r}}−\boldsymbol{{r}}_{{i}} \right) \\ $$$$\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{{r}}−\boldsymbol{{r}}'\mid}\right)=−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right) \\ $$$$\phi\left(\boldsymbol{\mathrm{r}}\right)=\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mathrm{4}\pi\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}' \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}'\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}\right){dV}' \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\left(−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right)\right)\mathrm{d}{V}' \\ $$$$=−\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right){dV}' \\ $$$$=−\rho\left(\boldsymbol{\mathrm{r}}\right) \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=−\rho \\ $$

Question Number 222924    Answers: 2   Comments: 0

Question Number 222922    Answers: 1   Comments: 0

Prove:∫_0 ^1 J_0 (ln(1/x))dx=((√2)/2)

$$ \\ $$$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} {J}_{\mathrm{0}} \left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right){dx}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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