How do I integrate e^(3x)?
546126
intrgrate o e^(3x)=(1/3)e^(3x)
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
769357
It can be helpful here to express this in terms of complex exponentiation. Note that sin x = (e^(ix) - e^(-ix))/(2i) and cos x = (e^(ix) + e^(-ix))/2. So:\n\n(sin x)^3 * (cos x)^4\n(e^(ix) - e^(-ix))^3/(2i)^3 * (e^(ix) + e^(-ix))^4/2^4\n(e^(3ix) - 3e^(ix) + 3e^(-ix) - e^(-3ix))/(-8i) * (e^(4ix) + 4e^(2ix) + 6 + 4e^(-2ix) + e^(-4ix))/16\n-1/(128i) * (e^(7ix) - e^(-7ix) + e^(5ix) - e^(-5ix) - 3e^(3ix) + 3e^(-3ix) - 3e^(ix) + 3e^(-ix))\n-1/64 (sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x))\n\nThus, all that tedious mucking about in the complex numbers has allowed us to reduce a hard-to-integrate product of powers of trigonometric functions into a sum of very to simple integrate functions. Thus:\n\n∫sin^3 x cos^4x dx\n-1/64∫sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x) dx\n-1/64 (- cos (7x)/7 - cos (5x)/5 + cos (3x) + 3 cos x) + C\ncos (7x)/448 + cos (5x)/320 - cos (3x)/64 - 3 cos x/64 + C
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
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
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
885310
rt2logx+C\n\n2.sectheta+C\n\n3.is it [3^(2x)+1/x^(1/3)]dx?
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
944695
12 * (3-e^-3x)^3 *  e^-3x
677393
it is probably easiest to expand into a polynomial like bboy does. or use u substitution:\nlet u =1-3x\ndu=-3 dx\nso integral of -2/3 u^2 = -2/9u^3 = -2/9(1-3x)^3
920193
e^(xlnx )=x^x\nderivative is  1+ln x
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.
643522
The derivative of e^[f(x)] is f'(x)e^[f(x)]\nThat is, it is the derivative of the exponent times e raised to the original exponent.\ny=3xe^(5x)\nYou have to use the product rule.\ny' = (3x)'e^(5x) + (3x)(e^(5x))\n= 3e^(5x) + (3x)(5)(3^(5x))\n= 3e^(5x) + 15xe^(5x)\n\ny=(9x+5)^(3/5)+e^(3/5)\ny' = (3/5)((9x+5)^(-2/5))(9) + 0\n= (27/5)(9x+5)^(-2/5)\nsince there was no x involved in e^(3/5), then e^(3/5) is just a constant (and the derivative of a constant is 0)\nRemember: When taking the derivative of e, the exponent NEVER changes.
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
971343
neither\nit is -1/2 e^cos 2x + C\n\nto confirm differentiate the result and find itr\n\nif you need to integate let cos 2x = y\nthen -2 sin 2x dx = dy\nsin 2x dx = -dy/2\n\nput the value we need to integrate - e^y dy/2 \nresult = - e^y/2 + C\n       = -1/2e^(cos 2x) + C
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.