evaluate the integral: (e^3x-4cosx)dx?
108152
(1/3)*e^(3x)-4sinx + c, where c is any arbitrary constant.
794871
assuming 0 and 3 are limits of integration:\nv(t) = dx/dt = t^2 - 2t\nx = (1/3)t^3 - t^2 + C\nx = (1/3)3^3 - 3^2 - 0 + 0\nx = 9 - 9 = 0\n\nIf the limits are 0 and 2, \nx = 8/3 - 4 = -4/3\n\nGoing backwards,\n(1/3)t^3 - t^2 = 8/3\nt^3 - 3t^2 -8 = 0\nt ≈ 3.612887865
278447
answer No 1 is bullshit. Answer No 2, I couldn't even understand what he/she wants to say.\nBut answer 3 is correct:\nint(e^(x-y)=int(e^-y e^x) = e^-y int(e^x) = e^-y e^x +c=\ne^(x-y) +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
571352
I am assuming I means intergral, the way you find the areaunder a curve.\n\nA = Integral of (x^3+4x-3.6) = x^4/4 + 2x^2-3.6x\n\nusing the formula Integral of (x^n) = x^(n-1)/n\n\nOf course you need limits of integration. i.e. between x=2 & x=5 to get a numerical answer. i.e. 23.5\n\n\nUse these already proven exponent rules.\n\na^0 = 1,    so 5^0 = 1 \n\na^-n = 1/a^n,    so 5^-1 = 1/5^1 = 1/5 or 0.2\n\n\nAlso:\n5^0 = 1 because of the following exponent rule & division\n\na^m/a^n = a^(m-n) \n\nso 5^3/5^3 = 5^(3-3)= 5^0 \n\nhowever  \n\nso 5^3/5^3 = 125/125 = 1\n\nTherefore\n\n5^0 = 1\n\nThe exception to the rule is 0^0, because you can't divide by zero, although wriiten as a function its limit approcahes 0. But that's another class.\n\nYou can use the same process to come to the answer for\n\n5^-1 \n\n5^2/5^3 = 5^-1 = 25/125 = 1/5 or 0.2\n\nThese rules however are already proven.\n\na^0 = 1\n\na^-n = 1/a^n
796927
try the link below...... hope it can put an end to ur integral problems
944695
12 * (3-e^-3x)^3 *  e^-3x
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
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
137569
(3/4P_ + 1/4pp) x (3/4S_ + 1/4ss)---maybe
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
933478
Using the Wolfram Integrator:\nhttp://integrals.wolfram.com/index.jsp\n\nOne obtains\n1/10 sec(2x)tan^4(2x) + 2/15 sec(2x)tan^2(2x) + 4/15 tan(2x)\n\nUsing the other poster's logic - you get (u^2 + 1)^3
617827
3x^3y^2z^4\n\nbecause you divide the exponents by 4, but do the fourth root of the coefficient 81
798015
y=e^x - 1 cuts the x axis at x = 1 Thus the integral needs to be done in two parts as area is a scalar quantity not a vector.\n\nA = |∫(x = -3 to 0) (e^x - 1) dx| + |∫(x = 0 to 3)  (e^x - 1) dx|\n\n= | [e^x - x] (x = -3 to 0) | + | [e^x - x] (x = 0 to 3) |\n\n= |(e^0 - 0) - (e^(-3) - -3)| + |(e^(3) - 3) - (e^0 - 0)| \n\n= 2 + e^(-3) + e^3 - 4\n\n= e^3 + e^(-3) - 2\n\n≈ 18.14 units²
815664
Imaginary part of\n(e^(i*x)+ e^(2*i*x) + ...)\n=Imag(e^(i*x)/(1- e^(i*x))\n=Imag(e^(i*x)/\n  (1-cos(x)-i*sin(x)))\n=Imag(e^(i*x)\n   (1-cos(x)+i*sin(x))/\n   ((1-cos(x))^2+(sin(x)^2))\n=((cos(x))\n(sin(x))+(sin(x))\n(1-cos(x)))/(2-2*cos(x)\n=sin(x)/(2(1-cos(x)))
10077
e^(ix)=cos(x)+i*sin(x)\n\nx^i=e(i*ln(x))\n\nx^i=cos(ln(x))+i*sin(ln(x))