Evaluate the integral. Use Integration by Parts.?
954798
The integration by parts formula is:\nintegral(f'g) = fg - integral(fg').\nIf we make f' = sin2t and g = t, then that second integral will only involve sin, and it should make everything easier.\nSo, that means f = -0.5cos2t, and g = t. Then g' = 1, so we get:\nintegral(t sin 2t) = -0.5tcos2t - integral(-0.5cos2t * 1)\nThat last integral now just becomes -0.25sin2t.\nSo the answer is -0.5tcos 2t + 0.25sin2t (+c).
421543
use substitution:\n\nlet u=-x^2, then du= -2x dx, so 4x dx= -2du \n\nso now we have the integral of -2e^u wrt to u\nwhich is:\n\n-2e^u\n\nnow putting back in that u=-x^2\n\nwe have the answer:\n\n-2e^(-x^2)
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
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
989233
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.
963003
int(y+2/2y^2+3y+1 dy)\n\n1/2*y^2 - 1/y + 3/2*y^2 + y + c\n\n2*y^2 - 1/y + y + c
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
245944
The integral of a function over a definite interval.
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.
774662
Set u = Arccos(x) such that dx = sinu du.  Then the integral becomes the integral of (u^2)sinu du.  Now the integral of this is (by parts) = -(u^2)sinu + 2*(the integral of) (usinu du).  And the integral of usinu (by parts) = -ucosu+(the integral of)(cosu du).\n\nHope this helps,\n\nSteve
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.
806158
Distribute the -1/2 through the parentheses and you get a \n-1/4 integral e^2x sin x\n\nAdd this to both sides and you will get the 5/4 integral of e^2x sinx
727341
first integrate\n\nF(x) = x^3 -x^2  from [1, 3]\n= 3^3 -3^2 -1^3 + 1^2\n= 27 - 9 -1 + 1\n= 18\n\naverage = 18/(3-1) \n= 18/2\n= 9
739570
d2 - =int(c2/10)\n\neasy
957499
The average value formula in the interval (a,b) is\n\nAV = 1/(b-a) * DEFINT f(x)  {evaluated from a to b}\nAV = 1/(b-a) * {INT f(4) - INT f(-4)} \n\nThe integral will be 1/2(x^4 - 5x^2).  If you evaluate this at x=4 and then at x=-4 and do the math it will be 0.\n\nSo the answer is 0.
850555
You can do it either way, but one way might be better depending on the limits of your integrals. \n\nIf you integrate with repect to x first, you do a substitution u=yx^2,\ndu=2xy dx to get an anti-derivative of (1/2)e^(x^2 y). The limits of integration on x will determine how hard the next integral (with respect to y) will be.\n\nIf you integrate first with respect to y, you have to do an integration by parts with u=xy and dv=e^(x^2 y) dy. Then du=xdy and v=(1/x^2) e^(x^2 y). You will still have to integrate v du to get an anti-derivative of\n(y/x)e^(x^2 y) - (1/x^3) e^(x^2 y).\nAgain, the limits of your integration will dictate how difficult the next integral will be.\n\nIf you have to change the order of integration, BE SURE YOU GET THE LIMITS OF INTEGRATION RIGHT.
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards