INTEGRATION PROBLEM (see below?
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.
402001
Watch it ….! \n\ndelta\n\n x =t^2  ---> t= sqrt(x)\ndx = 2t dt .............(not dx = dt /2t )\n-->\ndt = dx/(2 Sqrt(x)\n\nand this Integral can not be done using Analytical Method; Also, the (method of Integration by Part) can not apply in such integral because either of the cos(t) and sqrt(t) have infinitive derivatives.
958105
Hi,\n\nThis not just ...sqrt(pi) as one respondent thought.\n\nWhat you need are two very useful tools for integration, part one is solved by changing variables and part two is solved with integration by parts.\n\n1. let x= (y-a)/sqrt(b), so dy/dx = sqrt(b) \n=> dx = (1/sqrt(b)) dy (this isn't strictly correct but it works here)\n\nso:\nI = int(exp(-(x-a)^2/b)dx,-inf..+inf) \nI = int(exp(-((x-a)/sqrt(b))^2)dx,-inf..+inf) \nI = int(exp(-y^2)*(1/sqrt(b))dy,-inf..+inf)\nI=(1/sqrt(b))*int(exp(-y^2)dy,-inf..+inf)\nI=(1/sqrt(b))*sqrt(pi)  (from our given result)\nI = sqrt(pi/b)\n\nThe seond part requries a decent understanding of integration by parts, and if you email me I can go through this with you.
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
888834
Integrate from -π/2 to π/2 and double it
904086
I would use substitution method on this one...\n\nFirst, break it up into the sum of 2 integrals...\n\nInt(cost/(1+sin^2t))dt + Int(-2sint/(1+cos^2t))dt\n\nFor the first integral let u = sint\n\ndu = cost\n\nThen you have an integral that you can evaluate ...\n\nInt(1/(1 + u^2)du = arctan(u)\n\nSubstitute cos(t) back in for u and you get arctan(cos(t)), evaluate that from 0 to pi/2.\n\nFollow the same reasoning for the 2nd one!
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
1183660
Use integration by parts.\nu = 4x, du = 4dx\ndv = e^-xdx, v = -e^-x\n\nint(udv) = uv - int(vdu), so\nint(4xe^-x)dx = -4xe^-x - int(-e^-x(4dx) + c\n= -4xe^-x + 4int(e^-x)dx + c\n= -4xe^-x - 4e^-x + c\n= -4e^-x(x+1) + c
807024
You need to use integration by parts for this one as well.\nLet's call the integral A; that is, \nA = integral [(e^-x)cos(2x) dx]\nLet u = cos(2x) and dv = e^-x dx, then\ndu = -2 sin(2x) and v = -e^-x\nA = uv - integral [v du]\nA = -(e^-x)cos(2x) - integral [2(e^-x)sin(2x) dx]\nNow we need to use integration by parts again.\nLet u = 2sin(2x) and dv = e^-x dx, then\ndu = 4cos(2x) and v = -e^-x dx\nuv - integral(v du)\n= -2(e^-x)sin(2x) + integral [4(e^-x)cos(2x) dx]\n= -2(e^-x)sin(2x) + 4*integral [(e^-x)cos(2x) dx]\nSo we have that \nA = -(e^-x)cos(2x) - integral [2(e^-x)sin(2x) dx]\nA = -(e^-x)cos(2x) - [-2(e^-x)sin(2x) + 4*integral [(e^-x)cos(2x) dx]]\nA = -(e^-x)cos(2x) + 2(e^-x)sin(2x) - 4*integral [(e^-x)cos(2x) dx]]\nNotice that the last integral is actually the integral we started with.\nA = -(e^-x)cos(2x) + 2(e^-x)sin(2x) - 4A\nSolve for A by adding 4A to each side\n5A = -(e^-x)cos(2x) + 2(e^-x)sin(2x)\nMulitply both sides by (1/5)\nA = (1/5)[-(e^-x)cos(2x) + 2(e^-x)sin(2x)]\nA = (1/5)(e^-x) [2sin(2x) - cos(2x)]
147855
Try http://integrals.wolfram.com/
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)