whats the integral of cos(t^2)?
402001
Watch it ….! \n\ndelta\n\n x =t^2  ---> t= sqrt(x)\ndx = 2t dt .............(not dx = dt /2t )\n-->\ndt = dx/(2 Sqrt(x)\n\nand this Integral can not be done using Analytical Method; Also, the (method of Integration by Part) can not apply in such integral because either of the cos(t) and sqrt(t) have infinitive derivatives.
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
465086
If cos(x^2-9) could be integrated exactly, then we'd be able to integrate cos(x^2) and i sin(x^2) so that we'd be able to integrate [cos(x^2) + i sin(x^2)] = e^(ix^2) and hence also e^(-x^2).  However, e^(-x^2) is the bell curve, which is not integrable using elementary functions.
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
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
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.
732315
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
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
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
747716
From the law of cosines,\nL^2 = R^2 + R^2 -2RRcos(π/3)\nL^2 = 2R^2 -2R^2cos(π/3)\nL^2 = 2R^2 + R^2 = 3R^2\nL = R√3
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
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
626687
r(t) = 1+cos[t]\n\nPolar Conversion:\n x= cos[t]\n y= sin[t]\n\nm = weight of the cardioid\n\nxg = (1/m)∫x*1ds\nyg = (1/m)∫y*1ds\n\nds = sqrt[(r')^2 + (r^2)*((t')^2)]dt\n\n(ds)^2 = [(r')^2 + (r^2)*((t')^2)](dt^2)\n\n(ds)^2 = (dr(t))^2 + ((r(t))^2)*(dt^2)\n\n \n(ds)^2 = [(Sin[t])^2 + 1 + (Cos[t])^2](dt)^2\n\n(ds)^2 = 4*[(Cos[t/2])^2]*(dt)^2\n\nds = 2*Cos[t/2]*dt\n\nor m = ∫ds = ∫2*Cos[t/2]*dt = 2*sqrt[2]. {0<t<(pi/2)}\n\n\nxg = [1/(2*sqrt[2])]∫xds \n\nxg = [1/(2*sqrt[2])]∫(1 + Cos[t])Cos[t](2Cos[t/2])dt  {0<t<(pi/2)}\n\nxg = 6/5  ->  1.2\n\n\nsame for yg, but:\n\nyg = [1/(2*sqrt[2])]∫(1 + Cos[t])Sin[t](2Cos[t/2])dt  {0<t<(pi/2)} \n\nyg = (-1 + 4*sqrt[2])/5  ->  0.93
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
729006
3sin(x) - 2[cos(x)]^2 = 0\n\nNow [cos(x)]^2 = 1 - [sin(x)]^2 from the rule [sin(x)]^2 + [cos(x)]^2 =1\n-->\n\n3sin(x) - 2{1 - [sin(x)]^2} = 0\n2[sin(x)]^2 + 3sin(x) - 2 = 0\n\nsubstitute sin(x) = t\n\n2t^2 + 3t - 2 = 0\n(2t -1) (t + 2) = 0\n\nSolving ...t = -2, t = 0.5\n\nSubstituting back for sin(x)\n\nSin(x) = -2, 0.5\n\nSin(x) = -2 is not in the range since -1<sin(x)<1\n\nTherefore sin(x) = 0.5\n\nx = 30 degrees, 150 degrees for 0<x<360 degrees
362556
MaX is right except for one thing:\nsqrt(cos^2(x)) is not equal to cos(x), but is equal to abs(cos(x)).\n\nHence the answer is:\nintegral(0,32pi/3, sqrt(2)*abs(cosx))\n=sqrt(2) * (21*integral(0,pi/2,cosx) + integral(pi/3,pi/2,cosx))\n= sqrt(2) * (21 + sin(pi/2) - sin(pi/3))\n= sqrt(2) * (21 + 1 - 0.5)\n= sqrt(2) * 21.5\n=30.40559...
1561
It is the (secant(x))^2; or you could write it as (cos(x))^(-2) or 1/(cos(x))^2.\n\nBut basically the secant squared is the most elegant answer.