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IntegrationQuestion and Answers: Page 111

Question Number 128471    Answers: 2   Comments: 0

... calculus ... Φ=^? ∫_0 ^( 1) (ln(x))^2 ln((√(−ln(x))) dx

...calculus...Φ=?01(ln(x))2ln(ln(x)dx

Question Number 128417    Answers: 1   Comments: 0

η = ∫_0 ^( 1) x^3 (1−x^3 )^(n−1) dx

η=01x3(1x3)n1dx

Question Number 128408    Answers: 1   Comments: 0

ρ = ∫ ((sin (4x))/(sin^4 (x)+cos^4 (x))) dx

ρ=sin(4x)sin4(x)+cos4(x)dx

Question Number 128385    Answers: 1   Comments: 0

∫_1 ^( π) determinant ((x^3 ,(lnx),(sinx)),((3x^2 ),(1/x),(cosx)),(6,(2x^(−3) ),(−cosx)))dx =?

1π|x3lnxsinx3x21xcosx62x3cosx|dx=?

Question Number 128373    Answers: 1   Comments: 0

∫_0 ^1 x^(3/2) (1−x)^(1/2) dx

01x32(1x)12dx

Question Number 128334    Answers: 1   Comments: 0

Ω = ∫ (x^2 /( (√((a+bx^2 )^5 )))) dx ; where : a; b >0

Ω=x2(a+bx2)5dx;where:a;b>0

Question Number 128346    Answers: 2   Comments: 0

Question Number 128316    Answers: 1   Comments: 0

nice calculus Ω= ∫_0 ^( ∞) ((sin^3 (x))/x^2 )dx=?

nicecalculusΩ=0sin3(x)x2dx=?

Question Number 128285    Answers: 1   Comments: 0

cos ((π/7))−cos (((2π)/7))+cos (((3π)/7)) =?

cos(π7)cos(2π7)+cos(3π7)=?

Question Number 128276    Answers: 1   Comments: 0

∫_e^2 ^( ∞) (dx/(x^3 ln x)) ?

e2dxx3lnx?

Question Number 128262    Answers: 2   Comments: 0

Question Number 128251    Answers: 1   Comments: 0

Question Number 128244    Answers: 0   Comments: 0

...nice calculus... prove that ::Ω= ∫_0 ^(π/4) ln(sin(x))d=((−π)/4)log(2)−(G/2) log(2sin(x))=Σ_(n=1) ^∞ ((−1)/n)cos(2nx) Ω= ∫_0 ^( (π/4)) {−log(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)}dx =((−π)/4)log(2)−Σ_(n=1) ^∞ ∫_0 ^( (π/4)) ((cos(2nx))/n)dx =((−π)/4)log(2)−Σ_(n=1) ^∞ [(1/(2n^2 ))sin(2nx)]_0 ^(π/4) =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ ((sin(((nπ)/2)))/n^2 ) =((−π)/4)log(2)−(1/2){(1/1^2 )−(1/3^2 ) +(1/5^2 )−..} =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) ∴ Ω =((−π)/4)log(2)−(G/2) ✓ G:= catalan constant...

...nicecalculus...provethat::Ω=0π4ln(sin(x))d=π4log(2)G2log(2sin(x))=n=11ncos(2nx)Ω=0π4{log(2)n=1cos(2nx)n}dx=π4log(2)n=10π4cos(2nx)ndx=π4log(2)n=1[12n2sin(2nx)]0π4=π4log(2)12n=1sin(nπ2)n2=π4log(2)12{112132+152..}=π4log(2)12n=1(1)n1(2n1)2Ω=π4log(2)G2G:=catalanconstant...

Question Number 128221    Answers: 1   Comments: 0

Given f(x+(1/x)) = x^4 −(1/x^4 )+2 then ∫_1 ^( 2) (1−x^(−2) )f(x)dx=

Givenf(x+1x)=x41x4+2then12(1x2)f(x)dx=

Question Number 128194    Answers: 1   Comments: 2

Question Number 128192    Answers: 0   Comments: 0

Question Number 128175    Answers: 1   Comments: 1

∫_0 ^( 1) ∫_0 ^( 1) ((dx dy)/(1−xy^3 )) ?

0101dxdy1xy3?

Question Number 128109    Answers: 1   Comments: 0

∅ = ∫ (x−2) (√((x+1)/(x−1))) dx

=(x2)x+1x1dx

Question Number 128082    Answers: 0   Comments: 0

β = ∫ ((1+ln (x))/(x.cos^2 (x))) dx

β=1+ln(x)x.cos2(x)dx

Question Number 128079    Answers: 1   Comments: 0

Ω = ∫ ln (x+(√(x^2 +a^2 )) )dx

Ω=ln(x+x2+a2)dx

Question Number 128073    Answers: 1   Comments: 0

∫ ((sin(2x))/((1 − x)^3 )) dx

sin(2x)(1x)3dx

Question Number 128057    Answers: 3   Comments: 0

∫_0 ^( 1) ∫_0 ^( 1) (1/(1−xy^2 )) dx dy =?

010111xy2dxdy=?

Question Number 128052    Answers: 3   Comments: 0

I = ∫ arctan (((x−2)/(x+2))) dx

I=arctan(x2x+2)dx

Question Number 128026    Answers: 0   Comments: 0

∫_(π/2) ^((2π)/3) (((arcsinhx)^2 )/(2cosx))dx

π22π3(arcsinhx)22cosxdx

Question Number 128025    Answers: 2   Comments: 0

Question Number 128023    Answers: 0   Comments: 0

....nice calculus...= Titu′s lemma:: for any positive numbers : a_1 ,a_2 ,...,a_n , b_1 ,b_2 ,...,b_n we have: (((a_1 +...+a_n )^2 )/(b_1 +...+b_n ))≤(a_1 ^2 /b_1 ) +...+(a_n ^2 /b_n ) proof : put : x=(x_1 ,...,x_n )∈R^n :y=(y_1 ,...,y_n )∈R^n (x.y)^2 ≤∣x∣^2 ∣y∣^2 (cauchy−schwarz inequality) (x_1 y_1 +...+x_n y_n )^2 ≤(x_(1 ) ^2 +...+x_n ^2 )(y_1 ^2 +...+y_(n ) ^2 ) by applying subsitution : x_i =(a_i /( (√b_i ))) , y_i =(√b_i ) (i=1,2 ,...,n) ((a_(1 ) ^2 +...+a_(n ) ^2 )/(b_2 +...+b_n ))≤(a_1 ^2 /b_1 )+...+(a_n ^2 /b_n ) ✓✓

....nicecalculus...=Tituslemma::foranypositivenumbers:a1,a2,...,an,b1,b2,...,bnwehave:(a1+...+an)2b1+...+bna12b1+...+an2bnproof:put:x=(x1,...,xn)Rn:y=(y1,...,yn)Rn(x.y)2⩽∣x2y2(cauchyschwarzinequality)(x1y1+...+xnyn)2(x12+...+xn2)(y12+...+yn2)byapplyingsubsitution:xi=aibi,yi=bi(i=1,2,...,n)a12+...+an2b2+...+bna12b1+...+an2bn

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