Calculating Integrals?
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
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
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
700700
try putting x^2 = i*t\nwhere i = complex iota\n t is varible \n\nthere integral will be in form : i/2*( cos(t)+i*sin(t) ) dt\n\nso answer will be : 1/2 * (i* sin (t) + i*cos(t) ) = 1/2 * e^ (i*t) = 1/2 * e^x^2\n\nThat's the answer....... :D\n\n\n[Assuming u know that e^(i*t) = cos(t)+i*sin(t) ]\n\ncan be solved easily ......... try it
888834
Integrate from -π/2 to π/2 and double it
979363
You can't do this integral straight-up because it is a special case.  The integral is:\n\nsqrt(pi/2)Fresnel(sqrt(2/pi)x)\n\nBelow are websites with an Integral Calculator and the Fresnel function\n\n-------\n\nHope this helps
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
739848
u can integrate by parts with u= ln p and dv = p^5 or v= p^6/6\n\n\nu = ln p and v = p^6/6 dv = p^5\n\nint udv = uv - int v du = p^6 ln p/6 - int (p^6/6/p)\n= p^6 ln p/ 6 - int p^5/6 = (p^6 lnp)/6 - p^6 + c
761146
You're trying to equate them.  What you really need to do is take their difference and then take the integral of that term:\n\nA = ∫(lnx - (x-2))dx = ∫lnx dx - ∫(x-2)dx\n\nLooks a little easier now, eh?\n\nThere's still a problem.  To find the limits of integration, you have to solve the 2 eqns simultaneously to get the intersection points.  I don't think there is an algebraic way to do that, so here are the limits (determined numerically); x = .158594 and x = 3.1465
101992
Here are some sites that have saved my behind.  :)\n\nhttp://integrals.wolfram.com/index.jsp\n\nhttp://www.calc101.com/webMathematica/integrals.jsp\n\nhttp://mcraefamily.com/MathHelp/CalculusTableOfIntegrals.htm
242020
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ 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.
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
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!
245944
The integral of a function over a definite interval.