how do you integrate this?
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
883869
http://integrals.wolfram.com/index.jsp\nhas a beaut integrator\n\nIntegrate[Log[Cos[x]], x] == \nI  2              (2 I) x\n- x  - x Log[1 + E       ] + \n2\n \n  x Log[Cos[x]] + \n \n  I              (2 I) x\n  - PolyLog[2, -E       ]\n  2\n\nI tried to paste the devilish-looking solution
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)
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)]
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
905517
I answered this question for jc yesterday!  Search for questions to find the answer.\n\n\nJust integrate p(x) along the rod.
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.
94569
Your integral needs either limits (from 1 to X, for example) or a constant added to it. With limits, the apparent contradiction vanishes. With an arbitrary constant, it can be accounted for.
199351
I haven't worked with this in a long time and I'm not sure this will work, but try starting w/ this:\n1) Define an equation that describes the " ∞ " with polar coordinates (something like r = C1 - C2*theta);\n2) If L is lengh, delta.L = delta.theta / (2 pi) * r\nand dL = r/(2 pi) dtheta  ;\n3) substitute for r and integrate from 0 to 2pi (or the necessary for one complete " ∞ ".\n \nProbably you'll find more than one mistakes here, but I think the general idea is correct.
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.
806158
Distribute the -1/2 through the parentheses and you get a \n-1/4 integral e^2x sin x\n\nAdd this to both sides and you will get the 5/4 integral of e^2x sinx
1156022
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out. \nIf you get stuck, send me an e-mail or add additional details.
83265
That's an elliptic integral.  Check out the link.
1136463
It's been a while since I had calculus, but I think it's something like this:\n\nYou'd have to take the integral from 0 to k (treat k as a constant). When you have that, take that value (which is equivalent to the area) and set it equal to 3. Then, algebraically, solve for k.
38987
See the link below. It is much too detailed to go into here. This will give you instructions on how to do it.
950010
replace [Cos(2x)]^2 with 1 - [Sin(2x)]^2.\n\nnow multiply to get...[Sin(2x)]^3 - [Sin(2x)]^5.\n\nthen separate integral into two parts...\n\nnow use power rule and chain rule and don't forget to add a constant.