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Question Number 122748    Answers: 3   Comments: 0

∫_0 ^3 (dx/((3−x)(√(x^2 +1)))) ?

$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{{dx}}{\left(\mathrm{3}−{x}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:? \\ $$

Question Number 122734    Answers: 1   Comments: 0

Question Number 122713    Answers: 0   Comments: 0

... advanced math ... two simple and nice integrals: prove that:: Ω_1 =∫_0 ^( ∞) ((sin(e^(−γ) x)ln(x))/x) dx=0 Ω_2 =∫_0 ^( ∞) ((sin(x^((√2)/2) )ln(x))/x)dx=−πγ note :: γ : Euler−mascheroni constant. .m.n.

$$\:\:\:\:\:\:\:\:\:\:\:\:...\:\:{advanced}\:\:{math}\:... \\ $$$$\:\:\:\:\:{two}\:\:{simple}\:{and}\:{nice}\:{integrals}: \\ $$$$\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({e}^{−\gamma} {x}\right){ln}\left({x}\right)}{{x}}\:{dx}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega_{\mathrm{2}} \:=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \right){ln}\left({x}\right)}{{x}}{dx}=−\pi\gamma\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{note}\:::\:\gamma\::\:\:\mathscr{E}{uler}−{mascheroni} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{constant}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 122697    Answers: 3   Comments: 0

∫ (dx/( (√((x−a)(b−x))))) ?

$$\:\:\int\:\frac{{dx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}}\:? \\ $$

Question Number 122689    Answers: 1   Comments: 0

Evaluate the integral ∫ (((x)^(1/3) +1)/( (x)^(1/3) −1)) dx

$$\:{Evaluate}\:{the}\:{integral}\: \\ $$$$\:\:\int\:\frac{\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{1}}\:{dx}\: \\ $$

Question Number 122671    Answers: 1   Comments: 2

...nice integral... prove that :: ∫_0 ^( ∞) cos(πnx)((1/x^2 )−((πcoth(πx))/x))dx =^(???) πln(1−e^(−πn) ) .m.n.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{integral}... \\ $$$$\:\:\:{prove}\:{that}\:\::: \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {cos}\left(\pi{nx}\right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\pi{coth}\left(\pi{x}\right)}{{x}}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{???} {=}\pi{ln}\left(\mathrm{1}−{e}^{−\pi{n}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$

Question Number 122636    Answers: 1   Comments: 0

...nice calculus... In AB^Δ C prove :: ∗: sin((A/2))sin((B/2))sin((C/2))≤(1/8) ......................... ∗∗:: max(cos((A/2))cos((B/2))cos((C/2)))=?

$$\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:{prove}\:::\: \\ $$$$\:\:\:\ast:\:\:{sin}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{B}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$......................... \\ $$$$\:\:\:\:\ast\ast::\:\:\:{max}\left({cos}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{B}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\right)=? \\ $$$$\:\:\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 122631    Answers: 2   Comments: 2

solve ∫_((a−1)^2 ) ^a^2 cosh^(−1) (1/( (√(a−(√x))))) dx with a>0

$$\mathrm{solve}\:\underset{\left({a}−\mathrm{1}\right)^{\mathrm{2}} } {\overset{{a}^{\mathrm{2}} } {\int}}\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:{dx}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$

Question Number 122626    Answers: 2   Comments: 0

Question Number 122625    Answers: 0   Comments: 0

... advanced integral... i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=γ ii: ψ(x)=∫_0 ^( ∞) ((e^(−t) /t) −(e^(−tx) /(1−e^(−t) )))dt solution :{_(2 : ln(n) =^(easy) ∫_0 ^( 1) ((x^(n−1) −1)/(ln(x)))dx (∗∗)) ^(1: H_n = Σ_(k=1) ^n (1/k) =∫_0 ^( 1) ((1−x^n )/(1−x)) dx (∗)) (∗)−(∗∗): H_n −ln(n)=∫_0 ^1 (((1−x^n )/(1−x)) −((x^(n−1) −1)/(ln(x))))dx lim_(n→∞) (x^n )=^(0<x<1) 0 lim_(n→∞) (H_n −ln(n))=∫_0 ^( 1) ((1/(1n(x)))+(1/(1−x)))dx γ= ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx ✓ ............................. ψ(x)=^(easy) −γ+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt ψ(x)=−∫_0 ^( 1) (1/(ln(t)))+(1/(1−t))dt+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt =∫_0 ^( 1) −(1/(ln(t))) +((1−t^(x−1) −1)/(1−t))dt =−∫_0 ^( 1) (1/(ln(t)))+(t^(x−1) /(1−t)) dt=^(t=e^(−y) ) =−∫_∞ ^( 0) ((1/(−y))+(e^(−yx+y) /(1−e^(−y) )))(−e^(−y) )dy =∫_0 ^( ∞) (e^(−y) /y)−(e^(−yx) /(1−e^(−y) ))dy ∵ ψ(x)=∫_0 ^( ∞) ((e^(−y) /y)−(e^(−yx) /(1−e^(−y) )))dy ✓

$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:\:{integral}... \\ $$$$\:\:\:\:\:{i}:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{ln}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}=\gamma \\ $$$$\:\:\:\:\:\:{ii}:\:\psi\left({x}\right)=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{e}^{−{t}} }{{t}}\:−\frac{{e}^{−{tx}} }{\mathrm{1}−{e}^{−{t}} }\right){dt} \\ $$$$\:\:\:\:{solution}\::\left\{_{\mathrm{2}\::\:{ln}\left({n}\right)\:\overset{{easy}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{n}−\mathrm{1}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:\:\:\left(\ast\ast\right)} ^{\mathrm{1}:\:\:{H}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:{dx}\:\:\left(\ast\right)} \right. \\ $$$$\:\:\left(\ast\right)−\left(\ast\ast\right):\:\:{H}_{{n}} −{ln}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:−\frac{{x}^{{n}−\mathrm{1}} −\mathrm{1}}{{ln}\left({x}\right)}\right){dx} \\ $$$$\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \left({x}^{{n}} \right)\overset{\mathrm{0}<{x}<\mathrm{1}} {=}\mathrm{0} \\ $$$$\:\:{lim}_{{n}\rightarrow\infty} \left({H}_{{n}} −{ln}\left({n}\right)\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}{n}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx} \\ $$$$\:\:\gamma=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{ln}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}\:\:\checkmark \\ $$$$............................. \\ $$$$\:\:\:\:\psi\left({x}\right)\overset{{easy}} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\psi\left({x}\right)=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({t}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{{ln}\left({t}\right)}\:+\frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} −\mathrm{1}}{\mathrm{1}−{t}}{dt}\: \\ $$$$\:\:\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({t}\right)}+\frac{{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}\:{dt}\overset{{t}={e}^{−{y}} } {=} \\ $$$$\:\:\:\:\:=−\int_{\infty} ^{\:\mathrm{0}} \left(\frac{\mathrm{1}}{−{y}}+\frac{{e}^{−{yx}+{y}} }{\mathrm{1}−{e}^{−{y}} }\right)\left(−{e}^{−{y}} \right){dy} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{y}} }{{y}}−\frac{{e}^{−{yx}} }{\mathrm{1}−{e}^{−{y}} }{dy}\: \\ $$$$\because\:\psi\left({x}\right)=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{e}^{−{y}} }{{y}}−\frac{{e}^{−{yx}} }{\mathrm{1}−{e}^{−{y}} }\right){dy}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 122614    Answers: 4   Comments: 0

Question Number 122588    Answers: 2   Comments: 0

Find the arc length of the curve x=(1/4)y^2 −(1/2)ln (y) between the points with the ordinates y=1 and y=2.

$${Find}\:{the}\:\mathrm{arc}\:{length}\:{of}\:{the}\:{curve}\: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({y}\right)\:{between}\:{the}\:{points} \\ $$$${with}\:{the}\:{ordinates}\:{y}=\mathrm{1}\:{and}\:{y}=\mathrm{2}. \\ $$$$ \\ $$

Question Number 122587    Answers: 5   Comments: 0

∫_1 ^(16) arctan (√((√x) −1)) dx ∫_0 ^(π/2) sin (2x) arctan (sin x) dx

$$\:\:\underset{\mathrm{1}} {\overset{\mathrm{16}} {\int}}\:\mathrm{arctan}\:\sqrt{\sqrt{{x}}\:−\mathrm{1}}\:{dx}\: \\ $$$$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:\mathrm{arctan}\:\left(\mathrm{sin}\:{x}\right)\:{dx}\: \\ $$

Question Number 122572    Answers: 1   Comments: 0

Question Number 122544    Answers: 0   Comments: 0

any one can explain me about Fresnel integral ?

$$\:{any}\:{one}\:{can}\:{explain}\:{me}\:{about} \\ $$$${Fresnel}\:{integral}\:? \\ $$

Question Number 122542    Answers: 1   Comments: 0

∫_0 ^3 (√(1+cos (x^2 ))) dx ?

$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\sqrt{\mathrm{1}+\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}\:{dx}\:? \\ $$

Question Number 122528    Answers: 1   Comments: 0

...nice calculus... Ω =∫_0 ^( (π/2)) xsin(x).cos(x)ln(sin(x).ln(cos(x))dx =^(???) (π/(16))−(π^3 /(192)) ✓

$$\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {xsin}\left({x}\right).{cos}\left({x}\right){ln}\left({sin}\left({x}\right).{ln}\left({cos}\left({x}\right)\right){dx}\right. \\ $$$$\overset{???} {=}\:\frac{\pi}{\mathrm{16}}−\frac{\pi^{\mathrm{3}} }{\mathrm{192}}\:\checkmark \\ $$

Question Number 122506    Answers: 1   Comments: 0

Question Number 122496    Answers: 2   Comments: 0

V=∫_0 ^3 ((x dx)/( (√(x+1))+(√(5x+1)))) T=∫_(−π/2) ^(π/2) (√(cos x−cos^3 x)) dx

$$\:\:{V}=\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{{x}\:{dx}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{5}{x}+\mathrm{1}}} \\ $$$$\:{T}=\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\sqrt{\mathrm{cos}\:{x}−\mathrm{cos}\:^{\mathrm{3}} {x}}\:{dx}\: \\ $$

Question Number 122469    Answers: 1   Comments: 5

Question Number 122461    Answers: 2   Comments: 2

∫ (x/((x^2 +a^2 )(x^3 +b^2 ))) ?

$$\:\int\:\frac{{x}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +{b}^{\mathrm{2}} \right)}\:? \\ $$

Question Number 122432    Answers: 0   Comments: 0

Question Number 122430    Answers: 1   Comments: 0

∫ (x^3 /( (√(4−x^2 ))+x^2 −4)) dx

$$\:\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} −\mathrm{4}}\:{dx}\: \\ $$

Question Number 122397    Answers: 4   Comments: 0

a\∫((2x+1)/(1+x))(√((1−x)/(1+x)))dx c\∫(dx/( (√x)+(x)^(1/3) )) b\∫(dx/(x+(√(x−1)))) d\∫(dx/( (1+x)(√(1+x+x^2 ))))

$$\mathrm{a}\backslash\int\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{1}+\mathrm{x}}\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx}\:\:\:\:\:\:\:\:\mathrm{c}\backslash\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}}+\sqrt[{\mathrm{3}}]{\mathrm{x}}} \\ $$$$\mathrm{b}\backslash\int\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}−\mathrm{1}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{d}\backslash\int\frac{\mathrm{dx}}{\:\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }} \\ $$

Question Number 122366    Answers: 1   Comments: 0

Question Number 122349    Answers: 1   Comments: 1

Find the polynomial P(x) of least degree that has a maximum equal to 6 at x=1 and minimum equal to 2 at x=3.

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{of}\:\mathrm{least}\:\mathrm{degree} \\ $$$$\mathrm{that}\:\mathrm{has}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{6}\:\mathrm{at}\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{minimum}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2}\:\mathrm{at}\:\mathrm{x}=\mathrm{3}.\: \\ $$

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