Question and Answers Forum

All Questions   Topic List

IntegrationQuestion and Answers: Page 34

Question Number 179175    Answers: 0   Comments: 2

Find ∫x^5 (√(x^3 +1)) dx Answer: I= (2/(45)) (3x^3 −2) (√((x^3 +1)^3 )) + c

Findx5x3+1dxAnswer:I=245(3x32)(x3+1)3+c

Question Number 179181    Answers: 3   Comments: 0

Help-me! Use double integral to find the area of the region bounded by the following curves given in the plane shown below: y^2 = 4x and x^2 = 4y

Helpme!Usedoubleintegraltofindtheareaoftheregionboundedbythefollowingcurvesgivenintheplaneshownbelow:y2=4xandx2=4y

Question Number 179107    Answers: 1   Comments: 0

I= ∫ (x^2 /(sin (2arctan (e^x )))) dx , Find I

I=x2sin(2arctan(ex))dx,FindI

Question Number 179066    Answers: 1   Comments: 0

If f(x)=∫ (x^2 /(x^2 +tan x)) dx then ∫ ((tan x)/(x^2 +tan x)) dx =?

Iff(x)=x2x2+tanxdxthentanxx2+tanxdx=?

Question Number 179042    Answers: 2   Comments: 0

Question Number 178988    Answers: 1   Comments: 0

C = ∫_0 ^(π/2) ((cos^2 x)/(4sin^2 x+cos^2 x)) dx =?

C=π/20cos2x4sin2x+cos2xdx=?

Question Number 178922    Answers: 1   Comments: 0

Question Number 178879    Answers: 0   Comments: 0

Question Number 178832    Answers: 4   Comments: 0

Question Number 178712    Answers: 1   Comments: 0

Question Number 178693    Answers: 1   Comments: 6

Prove that ∫_0 ^(π/2) ((tan 2x)/( (√(sin^4 x+4cos^2 x)) −(√(cos^4 x+4sin^2 x))))=1

Provethat0π2tan2xsin4x+4cos2xcos4x+4sin2x=1

Question Number 178692    Answers: 1   Comments: 0

∫((tan (ln x).tan (ln (x/2)).tan (ln 2))/x)dx

tan(lnx).tan(lnx2).tan(ln2)xdx

Question Number 178657    Answers: 0   Comments: 0

calculate I=∫_0 ^( (π/4)) (( sin(x))/(1+ tan(2x))) dx = ?

calculateI=0π4sin(x)1+tan(2x)dx=?

Question Number 178563    Answers: 2   Comments: 0

∫e^x^2 dx=?

ex2dx=?

Question Number 178550    Answers: 1   Comments: 0

Question Number 179991    Answers: 1   Comments: 0

Question Number 178341    Answers: 1   Comments: 0

Question Number 178246    Answers: 1   Comments: 0

Question Number 177913    Answers: 0   Comments: 0

if I_n =∫xsin^n x dx and I_n = −((xsin^(n−1) x cosx )/n) +((sin^n x )/n^2 )+f(n)I_(n−2) then f(n) = ?

ifIn=xsinnxdxandIn=xsinn1xcosxn+sinnxn2+f(n)In2thenf(n)=?

Question Number 177906    Answers: 1   Comments: 0

∫_0 ^1 ((sin x(cos^2 x−cos^2 (π/5))(cos^2 x−cos^2 ((2π)/5)))/(sin 5x)) dx =?

10sinx(cos2xcos2π5)(cos2xcos22π5)sin5xdx=?

Question Number 177820    Answers: 2   Comments: 0

∫_(π/2) ^π (dx/((sin x−2cos x)^2 )) =?

ππ/2dx(sinx2cosx)2=?

Question Number 177784    Answers: 1   Comments: 0

Question Number 177775    Answers: 1   Comments: 0

∫((sin2x)/( (√(cos^4 x+1))))dx

sin2xcos4x+1dx

Question Number 177604    Answers: 0   Comments: 0

Calculate 𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^? −G ∼ Solution ∼ 𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx = Σ_(k=0) ^∞ (−1 )^( k) ∫_0 ^( 1) x^( 2k) ln(x) dx =^(i.b.p) Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx } = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) )) = − G ( Catalan constant ) ∴ 𝛗 = − G ■ m.n

Calculateϕ=01ln(x)1+x2dx=?GSolutionϕ=01{k=0(1)kx2kln(x)}dx=k=0(1)k01x2kln(x)dx=i.b.pk=0(1)k{[x1+2k1+2kln(x)]0111+2k01x2kdx}=k=0(1)1+k(1+2k)2=G(Catalanconstant)ϕ=Gm.n

Question Number 177541    Answers: 0   Comments: 3

Question Number 178487    Answers: 3   Comments: 0

∫ (dx/(cot^3 x sin^7 x)) =?

dxcot3xsin7x=?

  Pg 29      Pg 30      Pg 31      Pg 32      Pg 33      Pg 34      Pg 35      Pg 36      Pg 37      Pg 38   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com