How do you solve this integral??
965420
You have to use trigonmetric substitution, which is by far, the hardest type of integration.  It also involves making diagrams of triangles, which I hope you will bear with me with this limited interface.\n\nSo we want to solve for Integral (1/(1+y^2))dy.  You have to use trigonometric substitution, and \n\nlet y = tan(x).  Then dy = (sec(x))^2, or "secant squared x", and we just directly substitute that.\n\nIntegral (1/(1+y^2))dy = \nIntegral (1/(1 + (tan(x))^2) (sec(x))^2 dx =\n\nNote:  In trigonometric substitution, unlike the regular type, you're also replacing dy for what you solved for.\n\nUsing the trig identity (sec(x))^2 = (tan(x))^2 + 1, we get\n\nIntegral (1/(sec(x))^2 * (sec(x))^2) dx = (the two values will cancel each other out into 1)\nIntegral (1)dx =\n\nx + C\n\nRemember that above, we let y = tan(x), so we have to solve for x, and we get x = tan(inverse)y, or tan(x) with a -1 value between the "tan" and the "x".  The reason for this step is we require an answer __in terms of y__.\n\nSo, our actual answer is tan(inverse)(y) + C
120241
You need to do a Poisson integral. Rather than try to write things out, I refer you to eqaution (3) on the mathworld site.\n\n\nhttp://mathworld.wolfram.com/PoissonIntegral.html\n\nI hope that helps.
748051
The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
759209
This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
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
414817
int = integral\nint[(sinx)^5dx]\nRewrite (sinx)^5 as sinx (sinx)^4\n(sinx)^4 = ((sinx)^2)^2\n(sinx)^2 = 1 - (cosx)^2\nThus, \n(sinx)^4 = ((sinx)^2)^2\n= (1-(cosx)^2)^2\nNow the integral is\nint[sinx(1-(cosx)^2)dx]\nUse substitution.\nLet u = cosx, then du = -sinxdx\nNow the integral is\nint[-(1-u^2)^2du] \nExpand (1-u^2)^2 and multiply through by the -1\n-(1-u^2)^2 = -(1 - 2u^2 + u^4) = -1 + 2u^2 - u^4 \nThe integral now equals\nint[(-1 + 2u^2 - u^4 )du]\n= -u +(2/3)u^3 - (1/5)u^5 + C, where C is a constant\nReplace u with cosx to get the final answer.\n-(cosx) +(2/3)(cosx)^3 - (1/5)(cosx)^5 + 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)
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.
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
837567
Interesting...there are actually two solutions. Let's use substitution:\n\nLet u = cos(x). du = -sin(x)dx.\n\nSo Integral[sin(x)cos(x)]dx\n\n= - Integral[udu]\n\n= -u^2/2\n\n= -[cos(x)]^2/2\n\nNow, let u = sin(x). du = cos(x)dx.\n\nSo Integral[sin(x)cos(x)dx]\n\n= Integral[udu]\n\n= u^2/2 = [sin(x)]^2/2\n\nJust to check, I did a definite integral on both results, and got the same value for both.\n\nPretty neat...
1106006
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.
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
1092276
Let u = 3 - x^2, then du = -2x dx\nso (-1/2) du = x dx\nNow the integral is\n(-1/2)(1/u^2) du\n= (-1/2)(u^(-2)) du\n= (-1/2)u^(-1)/(-1) + C\n= (1/2)(u^(-1)) + C\n= 1/(2u) + C\n= 1/(2(3-x^2)) + C\nThere is no need to solve for x = sqrt(-u+3).  The x in the integral cancels with the x from du.