I need to know how to solve the following integrate?
100694
f= {square root(y)}  with\ny = 460x - x^2  in which ^2 = squaire\nf'= 1/[2f) and y'= 460-2x\nI use the chain rule. (That's the name I learnt in Europe)\nThe wanted derivate to x is f ' y' =\n{(460-2x)/2}/{square root(460x - x^2)}.
32210
integral of sec(x) = ln(|sec(x) + tan(x)|)\n\n\nsec(x) = (sec(x)(tan(x) + sec(x))/(tan(x) + sec(x))\n\n= (sec(x)tan(x) + sec(x)^2)/(tan(x) + sec(x))\n\nif u = (tan(x) + sec(x)), then du = (sec(x)tan(x) + sec(x)^2) \n\nso the integral is the integral of du/u = ln(|u|)
1092276
Let u = 3 - x^2, then du = -2x dx\nso (-1/2) du = x dx\nNow the integral is\n(-1/2)(1/u^2) du\n= (-1/2)(u^(-2)) du\n= (-1/2)u^(-1)/(-1) + C\n= (1/2)(u^(-1)) + C\n= 1/(2u) + C\n= 1/(2(3-x^2)) + C\nThere is no need to solve for x = sqrt(-u+3).  The x in the integral cancels with the x from du.
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
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
147855
Try http://integrals.wolfram.com/
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.
815728
first this is not a PDE but ODE\n\nsimple integration would do
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.
798293
use integration by parts:\n\nu=x\ndv= cos(x) dx\n\ndu =dx\nv= sin(x)\n\nintegral of x cosx dx = xsinx - integral of sin(x) dx\n= x sinx + cos(x) + C\n[
793695
' the integral of 1/e^u du = integral of e^{-u}du\n=-e^{-u} + C
532184
One method is to integrate by parts twice. There is another way where you can use complex methods: \ncos2x = Re e^(2ix)\n[Re means the real part]\n\nIntegral e^(-x)cos2x dx\n= Integral e^(-x) Re e^(2ix) dx\n= Re Integral e^((2i-1)x) dx\n= Re e^((2i-1)x)) / (2i-1) + C\n= Re e^(-x) e^(2ix) / (2i-1) + C\n\nWrite the above in the standard A + iB and then you can find your answer. Good luck.
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
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
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
978230
Let u = t\ndu = dt\n\nLet dv = sin(2t)dt\nv = -1/2 cos(2t) \n\nintegral (u dv) = uv - integral( v du)\n\n= -(1/2)*t*cos(2t) - integral[-1/2 cos(2t) dt]\n\n= -(1/2)*t*cos(2t) + 1/2*1/2*sin(2t)\n\n= -(t/2)cos(2t) + (1/4)sin(2t)\n\nIt looks like you did the integration by parts correctly, but you didn't evaluate the second integral.\n\nKeep at it!
1106006
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.