integrate (t^2)(cos(1-t^3)) dt using substitution...and please explain...?
929743
∫t²cos(1 - t³) dt\n\nI note that d(1 - t)/dt = -3t² \nThis is set for substitution of 1 - t³\n\nLet u = 1 - t Thus du = -3t²dt ie  -1/3 . du = t²dt\n\nSo ∫t²cos(1 - t³) dt\n= ∫cos(1 - t³) t²dt\n= ∫cosu (-1/3) . du\n= -1/3 sinu + c\n= -1/3 sin (1 - t³) + c
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
735191
Writing this out in more detail:\n1/2 ∫cos (2x) dx\nu=2x, du=2 dx, dx=du/2\n1/2 ∫cos u/2 du\n1/4 ∫cos u du\n1/4 sin u + C\n1/4 sin (2x) + C\n\nThe u comes from the fact that you are integrating cos 2x, not cos x.\n\nOh, and in the second step, you should have ∫(1-cos (2x))/2 dx -- the way you wrote it literally means ∫1 - x (cos 2)/2 dx, which is absurd.
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
246551
integral of sine squared.\nso theres a conversion thing which says \nsin^2 x = (1-cos 2x)/2\nso if you use that then u have to integrate 1/2 - cos 2x/2\nwhich is just 1/2x - sin 2x / 4
883869
http://integrals.wolfram.com/index.jsp\nhas a beaut integrator\n\nIntegrate[Log[Cos[x]], x] == \nI  2              (2 I) x\n- x  - x Log[1 + E       ] + \n2\n \n  x Log[Cos[x]] + \n \n  I              (2 I) x\n  - PolyLog[2, -E       ]\n  2\n\nI tried to paste the devilish-looking solution
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.
978230
Let u = t\ndu = dt\n\nLet dv = sin(2t)dt\nv = -1/2 cos(2t) \n\nintegral (u dv) = uv - integral( v du)\n\n= -(1/2)*t*cos(2t) - integral[-1/2 cos(2t) dt]\n\n= -(1/2)*t*cos(2t) + 1/2*1/2*sin(2t)\n\n= -(t/2)cos(2t) + (1/4)sin(2t)\n\nIt looks like you did the integration by parts correctly, but you didn't evaluate the second integral.\n\nKeep at it!
744903
1)\n\ntan^2(x) = sec^2(x) - 1\n\nSo the integral of sec^2(x) + 4 = tan(x) + 4x\n\n2)\n\ncsc(x) = 1/sin(x)\n\nthe integral is then of 1+sin(x)/cos^2(x).\nlet u = cos(x) then du = -sinx dx.\nso the integral is x+sec(x).\n\n3)\n\nlet u = cos(2x+1) then du = -(1/2)sin(2x+1) dx.\n\nso the integral is of -(1/2)/u^2 = 2/u = (1/2)/cos(2x+1) = sec(2x+1)/2\n\nSteve
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
281867
Well I assume you are trying to work out x and I also assume that it's not cos [to the power of] (x^3) but Cos [multiptied by] (x^3)\n\nYou need to integrate to remove the d/dx:\n\nCos integrated is Sin and you then have to divide by the differental of what's in the brackets\n\n=> sin(x^3) / 3x^2
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
1159338
Integrate using the method of substitution based on the fact that the derivative of log(y) + C = 1/y\n\nlet u = log(y) then du/dy = 1/y. So du = (1/y) dy\n\nUsing these substitutions we have:\n\nInt 1/y*log(y) dy  = Int 1/log(y)*(1/y) dy\n\n= Int 1/u du  = log(u)+C = log(logy)+C, since u = log(y)\n\nThis is correct since from the chain rule:\n\n d/dy log(logy)+C= 1/log(y)*d(logy)/dy= 1/(y*logy)
127159
one correction.\n\ndv/dr = cos(s+2t)\nd^2v/dr^2 = 0
32210
integral of sec(x) = ln(|sec(x) + tan(x)|)\n\n\nsec(x) = (sec(x)(tan(x) + sec(x))/(tan(x) + sec(x))\n\n= (sec(x)tan(x) + sec(x)^2)/(tan(x) + sec(x))\n\nif u = (tan(x) + sec(x)), then du = (sec(x)tan(x) + sec(x)^2) \n\nso the integral is the integral of du/u = ln(|u|)
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
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......