Using integration by parts, show that integration x squared cos 2x dx with limits pi and 0 is equal to pi/2?
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
1051703
collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
147855
Try http://integrals.wolfram.com/
385118
the circumference divided by 2 x PI
860131
It's undefined at the endpoint of the interval (pi/2).\nFirst go ahead and integrate it like you normally would.\nintegral of sin(x)/sqrt(cosx) dx\nUse substitution.  Let u = cos x, then du = - sinx dx\n= integral of -du/ sqrt(u) \n= integral -u^(-1/2) du\n= -u^(-1/2 +1) / (-1/2 + 1) \n= -u^(1/2) / (1/2)\n= -2u^(1/2)\n= -2 sqrt(u)\nReplace u with cosx\n= -2 sqrt(cos x)\nNow plug in pi/2\n- 2 sqrt(cos pi/2)\n= -sqrt(0)\n= 0\nPlug in pi/4\n-2 sqrt(cos pi/4)\n= -2sqrt( sqrt(2) / 2)\nNow subtract them\n0 - (-2sqrt(sqrt(2) /2))\n= 2sqrt( sqrt(2) / 2)
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
6838
∫xSec(x²)²dx = .5Sin(x²)/Cos(x²)
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
400264
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
786513
put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
874788
cos ^-1 0 = pi/2\n1/2 cos^-1 0 = pi/4\n\nso tan(1/2 cos^-1(0)) = tan pi/4 = 1
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.
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
631241
the integration by part is very long and complicated\ni used it many times\nlet u =logsinx          dv=dx\n   du= cosx/sinx          =x\nxlogsinx-int{xcosx/sinx}\nint{xcosx/sinx}  by sbutitution\nlet u=sinx   x=sin^-1(x)\ndu=cosx dx dx=du/cosx\nyou will get int{sin^-1(u)/u} remmeber u=sinx\nby part\nlet k=1/u          dv=sin^-1(u)\n     -1/u^2        int{sin^-1(u)}by part\n\nint{sin^-1(u)}\nlet m=sin^-1(u)                dv=du\n     =1/sqt(1-u^2)                u du\n=usinx^-1(u)-int{u/spt(1-u^2)}\nint{u/spt(1-u^2)} int this by subitution\n\nthen u will get\nsinxsin^-1(x)+(1/2)(1-sin^2x)\n\nso the final ans\nxlogsinx-sinxsin^-1(x)\n+(1/2)(1-sin^2x)+c\n\nplz try to answer this by yourself ,i hope i gave you the idea,thankx
403929
Integration by parts is done by breaking down an integral so that\n\nS u dv = uv - S v du (where S is the integration symbol)\n\nFor example, if we were to integrate xcos x dx,\n\nlet u = x\ndu = dx\n\nthen dv = cos x (all the other stuff we didn't use for u)\nv = sin x (just integrate dv)\n\nand the formula will look like\n\nS xcos x dx = xsin x - S sin x dx\n= xsin x + cos x + C\n\nIt's kinda an art form or experience in picking which part of the integral to make u and dv - there's really no rule that says u = cos x.  Experience (or trial and error) will show you that that's probably not the better choice though .
1128475
If you were to integrate f(x) from 0 to pi/2\n\nit wold be  int(f,0,pi/2).
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Let 'I' denote the integral sign.\nWe will use the method of substitution.\nLet e^2x = z.\nThen taking differentials of both sides 2*e^2x*dx = dz or 2*z*dx = dz or dx = dz/(2*z).\n\nNow 1/((e^-2x)+( e^2x)) can be written as (e^2x)/[1 + (e^2x)^2]\n\nHence given integral = I[(e^2x)/{1 + (e^2x)^2}]dx\n= I[z/(1 + z^2)]{dz/(2*z)} [substituting z = e^2x]\n= 0.5*I[1/(1 + z^2)]dz\n= 0.5*tan-1(z) + C [tan-1(z) denotes tan inverse z]\nwhere C = arbitrary constant of integration\n= 0.5*tan-1(e^2x) + C