Question and Answers Forum |
IntegrationQuestion and Answers: Page 93 |
![]() |
π=β«_0 ^( 1) ((ln(x^2 +1))/x^2 )dx f(a)=β«_0 ^( 1) ((log(ax^2 +1))/x^2 )dx f β²(a)=β«_0 ^( 1) (x^2 /((ax^2 +1)x^2 ))dx=(1/a)β«_0 ^( 1) (dx/(x^2 +((1/( (βa))))^2 )) =((βa)/a)[tan^(β1) (x(βa) )]_0 ^1 =((βa)/a)tan^(β1) ((βa)) f(a)=^((βa) =u) β«2tan^(β1) (u)du =2{u.tan^(β1) (u)ββ«(u/(1+u^2 ))du}+C =2(βa) tan^(β1) ((βa) )βln(1+a)+C f(0)=0=0+CβC=0 f(1)=π=2((Ο/4))βln(2)=(Ο/2)βln(2) |
..........nice calculus......... suppose that::: Ο(p)=β«_0 ^( β) ((ln(1+x))/((p+x)^2 )) ...β find the value of:: β«^( 1) _0 ((Ο(p))/(1+p))dp=?... |
....advanced calculus.... π=β«_0 ^( (Ο/2)) x.(tan(x))^(1/2) dx=?? |
......nice calculus..... prove:: β«_0 ^( 1) (1/(1+ln^2 (x)))dx=β«_(0 ) ^( β) ((sin(x))/(1+x))dx |
f(x)=β«_(βΞ /4) ^(Ξ β«/4) e^(xtant) dt |
β«_0 ^((50Ο)/3) β£sinxβ£dx |
(a) Let I(Ξ±)=β«_0 ^β e^(β(xβ(Ξ±/x))^2 ) dx Show that it is legitimate to take the derivative of I(Ξ±) and also Iβ²(Ξ±)= 0. Then show that I(Ξ±)=((βΟ)/2). (b) Use (a) to prove β«_0 ^β e^(β(x^2 +Ξ±^2 x^(β2) )) dx=((βΟ)/2)e^(β2Ξ±) . |
![]() |
![]() |
β« (dx/(sin^6 x)) ? |
If Ξ±>0 and Ξ²>0, prove β«_0 ^β ((ln(Ξ±x))/(Ξ²^2 +x^2 ))dx=(Ο/(2Ξ²))ln(Ξ±Ξ²) |
calculate A_Ξ» =β«_0 ^β ((cos^4 x)/((x^2 +Ξ»^2 )^2 ))dx 2) find the value of β«_0 ^β ((cos^4 x)/((x^2 +3)^2 ))dx |
if f(x)=x^3 β3x+2 determine f^(β1) (x) and β« f^(β1) (nf(x))dx with n integr |
find β« ((arctan(2x))/(x+3))dx |
find U_n =β«_(1/n) ^n (1β(1/x^2 ))arctan(x+(1/x))dx and lim_(nββ) U_n |
find β«_0 ^β ((arctan(x^2 ))/(x^4 +1))dx |
calculate β«β«_([0,1]^2 ) e^(β(x^2 +y^2 )) arctan(x^2 +y^2 )dxdy |
calculate β« (dx/(x^n (β(x^2 β1)))) |
find β«_0 ^1 (x^a /(1βx))dx |
......sdvanced cslculus...... if xβR^+ and:: π(x)=β«_0 ^( x) ((e^t β1)/t)ln((x/t))dt then prove that :: Ξ¨=β«_0 ^( β) e^(βx) π(x)dx=ΞΆ(2) |
β« (dx/(sin x (β(cos x)))) =? |
lim_(xβ0) (x^(2021) /(xβln(Ξ£_(k=0) ^(2020) (x^k /(k!))))) =^? 2021! |
calculate β«_0 ^β (t^a /(1+t+t^2 ))dt study first the convergence (a real) |
please a generall Form for C(n) C(n)=(4/Ο^2 )Ξ£_(k=1) ^n (β1)^(kβ1) ΞΆ(2k)ΞΆ(2nβ2k) |
![]() |