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IntegrationQuestion and Answers: Page 120

Question Number 122927    Answers: 1   Comments: 1

Question Number 122922    Answers: 1   Comments: 0

∫_0 ^(π/4) ((cos x+sin x)/(16sin 2x+9)) dx

π/40cosx+sinx16sin2x+9dx

Question Number 122919    Answers: 2   Comments: 0

Question Number 122885    Answers: 3   Comments: 1

Question Number 122884    Answers: 1   Comments: 1

Question Number 122882    Answers: 1   Comments: 0

... nice calculus... prove that ::: Ω=∫_0 ^( 1) (((x^ϕ −1)/(ln(x))))^2 dx=(√5) ln(ϕ) .m.n.

...nicecalculus...provethat:::Ω=01(xφ1ln(x))2dx=5ln(φ).m.n.

Question Number 122877    Answers: 1   Comments: 1

∫ (sin^(−1) (x))^2 dx ?

(sin1(x))2dx?

Question Number 122875    Answers: 1   Comments: 0

∫(((x^2 +1)dx)/(x^4 +x^2 +1)) = ...

(x2+1)dxx4+x2+1=...

Question Number 122867    Answers: 2   Comments: 0

Question Number 122853    Answers: 0   Comments: 2

Question Number 122838    Answers: 1   Comments: 0

∫ (dx/( (((x−1)^3 (x+2)^5 ))^(1/4) )) ?

dx(x1)3(x+2)54?

Question Number 122828    Answers: 1   Comments: 0

Question Number 122823    Answers: 1   Comments: 0

Question Number 122822    Answers: 1   Comments: 0

Question Number 122807    Answers: 1   Comments: 1

Question Number 122751    Answers: 3   Comments: 0

∫_0 ^(1/(√2)) ((x sin^(−1) (x^2 ))/( (√(1−x^4 )))) dx ?

1/20xsin1(x2)1x4dx?

Question Number 122748    Answers: 3   Comments: 0

∫_0 ^3 (dx/((3−x)(√(x^2 +1)))) ?

30dx(3x)x2+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.

...advancedmath...twosimpleandniceintegrals:provethat::Ω1=0sin(eγx)ln(x)xdx=0Ω2=0sin(x22)ln(x)xdx=πγnote::γ:Eulermascheroniconstant..m.n.

Question Number 122697    Answers: 3   Comments: 0

∫ (dx/( (√((x−a)(b−x))))) ?

dx(xa)(bx)?

Question Number 122689    Answers: 1   Comments: 0

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

Evaluatetheintegralx3+1x31dx

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.

...niceintegral...provethat::0cos(πnx)(1x2πcoth(πx)x)dx=???πln(1eπn).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)))=?

...nicecalculus...InABCΔprove:::sin(A2)sin(B2)sin(C2)18.........................::max(cos(A2)cos(B2)cos(C2))=?

Question Number 122631    Answers: 2   Comments: 2

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

solvea2(a1)2cosh11axdxwitha>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 ✓

...advancedintegral...i:01(1ln(x)+11x)dx=γii:ψ(x)=0(ettetx1et)dtsolution:{2:ln(n)=easy01xn11ln(x)dx()1:Hn=nk=11k=011xn1xdx()()():Hnln(n)=01(1xn1xxn11ln(x))dxlimn(xn)=0<x<10limn(Hnln(n))=01(11n(x)+11x)dxγ=01(1ln(x)+11x)dx.............................ψ(x)=easyγ+011tx11tdtψ(x)=011ln(t)+11tdt+011tx11tdt=011ln(t)+1tx111tdt=011ln(t)+tx11tdt=t=ey=0(1y+eyx+y1ey)(ey)dy=0eyyeyx1eydyψ(x)=0(eyyeyx1ey)dy

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