using a reduction formulA how do you find the integral(sec^3)2x*dx? I?
138355
First get rid of 2x by putting t = 2x\n( Read "eye 1" =  and "eye = " etc... capital letter I doesn't show up well on screen)\nI1 = Integral (sec^3 (t) dt/2)\nFor simplicity I will write Int for integral sign\nI = Int sec^3 t dt = Int sec t sec^2 t dt\nApply product rule\n\nI = sec t * tan t - Int sec t tan^2 t dt\nI = sec t tan t - Int sec t (sec^2 t - 1) dt\nExpand\nI = sec t tan t - Int sec^3 t dt + Int sec t dt\nBut second term is -I hence you have\n2I  = sec t tan t + Int sec t dt\n\nThis is standard form to evaluate using \n\nInt sec t = ln tan (pi/4 + t/2) + c\n I = 1/2 (sec t tan t + ln tan (pi/4 + t/2))+ C\nI1 = 1/2 I
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
904685
1)auxilary equation\n\nL^2+3L=0\n\n>>>>L=0,L= -3\n\ncomplementary function\n\nyc= C+De^(-3x)\n\n2)find a particular integral for\n                  d^2y/dx^2 +3dy/dx= x^2-2x\n\ntry y= ax^3+bx^2+cx...(1)\n\ndy/dx=3ax^2+2bx+c\n\nd^2y/dx^2= 6ax+2b\n\nsubstitute these into LHS of DE\n\n6ax+2b+3(3ax^2+2bx+c)=x^2-2x\n9ax^2+6(a+b)x +(2b+3c)=x^2-2x\n\nequating coefficients\n\n9a=1 >>>a= 1/9, 6(a+b)= -2\n         >>>b= -4/9 \n2b+3c=0 >>>>>>c=8/27\n\nsubstitute these values into (1)\n\nparticular integral\n\nyp=(1/9)x^3-(4/9)x^2+(8/27)x\n\n3)general solution\n\ny=yc+yp\n\n  =C+De^(-3x)\n    +(1/9)x^3-(4/9)x^2+(8/27)x\n\ni hope that this helps
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
969937
'\nu= sec x \ndu = tan x sec x dx\ntan x dx = du/sec x = du/u\ntherefore:\nintegral of tan x dx  = integral of du/u = ln |u| + c = ln| sec x| + c
541499
2x(x-5)-3(2x+1)\n2x^2-10x-6x-3\n2x^2-16x-3
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
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
801525
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
546126
intrgrate o e^(3x)=(1/3)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
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
809968
(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1\n\nSenx / cos x  - cosx/senx\n---------------------------------  + 2cos^2x\nSenx / cos x  + cosx/senx\n\n(sen^2 x - cos^2x) / senx cosx\n--------------------------------------  + 2cos^2 x\n(sen^2 x + cos^2x) / senx cosx\n\nThe fraction reduces on the above:\n\nsen^2 x - cos^2 x                       + 2 cos^2 x\n\nsen^2x + cos^2x = 1\n\nproved!
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.
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)