plz help, i have to answer this integral?
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
706536
Integral: ∫ [2/{(e^2x)+4}] dx\n\nFirst consider the expression: 2/{(e^2x) + 4}\n\n= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]\n\n= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]\n\n\nHence, ∫[2/{(e^2x) + 4}] dx\n\n=  ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx\n\n=  ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx\n\n=  (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx\n\nLet {e^(2x) + 4} = z. Then dz = 2e^(2x) dx\n\nHence the given integral becomes:\n(1/2) x - (1/4)∫(dz/z)\n\n= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration\n\nNow, substitute the value of 'z' in terms of 'x', thus the given integral becomes\n= (1/2) x - (1/4)ln|e^(2x) + 4| + c
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
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
147855
Try http://integrals.wolfram.com/
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
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.
245944
The integral of a function over a definite interval.
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
1128475
If you were to integrate f(x) from 0 to pi/2\n\nit wold be  int(f,0,pi/2).
283258
I=int dx/[(x^2)*(x^2+1)^.5]\n\nLet x=tan u\n   dx=secu^2 du\n\nSubtitute\n\nI=int (secu^2 du)/[(tanu^2)*(tanu^2+1)^.5]\n\nBut tanu^2+1=secu^2\n\nThen \nI=int (secu^2 du)/[(tanu^2)*(secu^2)^.5]\n--> \nI=int (secu^2 du)/[(tanu^2)*(secu)]\n-->\nI=int (secu du)/[(tanu^2)]\n\nBut\nsecu=1/cosu ..... and tanu=sinu/cosu\n\nThen\nI=int (1/cosu)/(sinu/cosu)^2\nI=int (1/cosu)*(cosu/sinu)^2\nI=int cosu/(sinu)^2\nI=int (cosu)*(sinu)^-2\nI= -(sinu)^-1\nI= -cscu + c .......... c is a constant\n\n\nBut tanu=x --> sinu=x/[(x^2+1)^0.5]\n\nthen\n-cscu= -1/sinu= -1/[x/(x^2+1)^0.5]= -[(x^2+1)^0.5]/x\n\nthen\nI = -[(x^2+1)^0.5] / x + c
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
226022
Let u = cot(x)\ndu = -csc^2(x)dx\n\nThis integral will then simplify to\n∫(-5*e^(u)du)\n     =-5*e^(u)\n     =-5*e^(cot(x))
403275
Look, to count this integral, you must make a subsitution:\n\nu = 1+5t\ndu = 5 dt\ndt = du/5\n\nint 8/(1+5t) dt =\n= int 8/u du/5 =\n= 8/5 * int (1/u) du =\n= 8/5 * ln u + C =\n= 8/5 * ln(1+5t) + 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...
905517
I answered this question for jc yesterday!  Search for questions to find the answer.\n\n\nJust integrate p(x) along the rod.
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.