Can anyone explain to me how I can find the antiderivative of cos(x^2)?
830695
UPDATE: Sorry for the edit, but on further review, it IS possible using an infinite summation. I can't really write it out here, so check this wikipedia link:\n\nhttp://en.wikipedia.org/wiki/Fresnel_integral\n\nYou can't do it with straight alegebra because there is no trigonometric function to expand f(x) = cos(x²), and there is no coefficient of x (so the chain rule doesn't apply). Basically, it's impossible with basic algebra.\n\nHowever, I put this function into Maple 7.0 and this is what it gave:\n\n1/2 * sqrt(2 * pi) * FresnelC(x * sqrt(2 / pi))\n\nSo apparently there exists some Fresnel function that can be used to derive the answer. I'm not sure though, that may be much more work than you want to do.
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
967981
y'=cos(x^2e^x)(2xe^x+x^2e^x)\n\ny''=-sin(x^2e^x)(2xe^x+x^2e^x)^2+\n      (2(e^x+xe^x)+(2xe^x+x^2e^x))cos(x^2e^x)
889651
It's just arctan (x^2).\n\nBy the fundamental theorem of calculus, if F(t) is the antiderivative of f(t) = arctan(t^2), then the integral of f(t) from 0.5 to x will give F(x) - F(0.5). The derivative of this function with respect to x then gives you f(x).
907483
1/x^2 = x^(-2)\nThe antiderivative is x^(-2+1)/(-2+1) + C\n= x^(-1)/-1 + C\n= -1/x + C\nIt is similar for 1/x^3\nYou do partial integrals the same way you do regular integrals.  You need to know which one you are integrating with respect to and then pretend the other variable(s) are constants.
750604
As Stuart has said above, you need to use the product rule to work this out.  The product rule states:\n\nthe derivative of f(x)g(x) is f(x)g'(x) + f'(x)g(x)\n\nSo, when you differentiate sinx wrtx you get cos and when you differentiate 2x wrtx you get 2.  Plugging these into the formula above you get:\n\n(f(x)g(x))'=2x(cosx)+2(sinx)\n                =2xcosx + 2sinx\n\nHope that this is helpful!
642641
please rewrite the formula in proper notation...i cannot tell where (x-10) is in relation to the argument of cos.\n\nis it y = 5 + 2*cos[pi/4*(x-10)]\nor y = 5 + [2*cos(pi/4)]*(x-10)??\n\nOk, here you go.  Since cos is 2pi periodic, then the argument will be 2pi periodic where, 0<pi/4*(x-10)<2pi, or\n0<x-10<8.  Which means for every time you add 8 radians to x, you will be at the exact location on the unit circle that you started.  Therefore, you are getting correct values for x (9.2936, 10.7064, etc.),but the question asks for the FIRST positive values.  Subtract 8 from each of your answers until you are at the lowest possible positive number.  (In this case you only subtract 8 once.)
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.
835948
function Cos^ (-1)x is an inverse of function Cos x\n(like square(x) and squareroot(x) act oppostite\nsquare(sqrootx) = x or sqroot(square(x) = x\nTherefore Cos[Cos^(-1)x = x\nNow to your question\nLet [Cos^(-squareroot3/2) = x\nTaking Cos on both side\nCos[Cos^(-1)(-squareroot3/2)] = Cos x\nor -squareroot3/2 = Cos x\nThis equation has many solutions as you can see (for every cycle of 360 deg you will have two solutions - one each in second quadrant and  third quadrant where Cosine (or cos) has negative values\nPrimarily we take the solution which between zero and 360 degrees\nTwo solutions are\nCos (180 -30) = Cos (180+30) = (- squareroot 3/2)\ntherefore 150 deg and 210 deg\nconverted to 150xpi/180 = 5/6 pi radians\nand 210xpi/180 = 7/6 pi radians\nSimilarly we can find a set of two solutions between 360 &720 deg, between 720 &1080 deg etc and so on\neg [360 + (180+30)] deg and [360 +180-30] deg [720 + (180+30)] deg [720+ (180-30)] deg and so on\nCheers\nSubhash
1020565
The derivative of the function given is\n\ny'=cos(2x)*2 - 2*cos(x) = 2(2cos^2(x)-1)-2*cos(x) \n= 4cos^2(x)-2cos(x)-2 = 2(2cos^2(x)-cos(x)-1)\n=2(2cos(x)+1)(cos(x)-1)\n\nFor y'=0, either 2cos(x)+1=0 or cos(x)-1=0, from which we get\n\ncos(x)=-1/2 (which gives either x=2*pi/3 + 2*pi*k, k an integer or x=-2*pi/3 + 2*pi*k, k an integer)\n\nor \ncos(x)=1, so x=2*pi*k, k an integer.
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
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
918747
For #1, there's no way you can solve that directly.  Note that it can be split up into three fractions, though.\n\n(x^2 + x + 1)/x = x^2/x + x/x + 1/x\n\nNow, we can actually reduce that, to\n\nx + 1 + 1/x\n\nWe solve for the antiderivative using the reverse power rule for derivatives.  That is, antiderivative for x^n is [x^(n+1)]/(n+1).  The exception is 1/x, where the antiderivative is ln|x|.\n\nSo the antiderivative of x+1+1/x is\n\n(x^2) / 2 + x + ln|x|\n\nBut we want the general antiderivative, so we add a constant C.\n\n(x^2)/2 + x + ln|x| + C
200165
I will assume cos(x)^3 means (cos x)^3 and does NOT mean  cos(x^3)\nFirst start by separating the integral into two pieces.\nI will use "int" to stand for the integral sign.\nint{(cosx)^3/(cosx)^2}dx - int{(sinx)^2/(cosx)^2}dx\n\nNow (cosx)^3/(cosx)^2 = cosx\nand (sinx)^2/(cosx)^2 = (tanx)^2 = (secx)^2 - 1\nThus, the integral is now equal to \n\nint{cosx}dx - int{(secx)^2 - 1}dx\n\nThe antiderivative of cosx is sinx.\nThe antiderivative of (secx)^2 is tanx.\nThe antiderivative of -1 is -x.\n\nHence, the solution is \nsinx - tanx + x + C\n\n(The signs changed for tanx and -x because there was a negative sign in front of their integral, and the negative sign has to be distributed).
170888
i am assuming we are differentiating with respect to x...\n\nd sin((k-x)^2) / dx = cos ((k-x)^2) * d (k-x)^2 / dx = cos((k-x)^2) * 2(k-x) * (-1)\nby Chain rule
647978
Since x=cos(cos(x)), then -1<= x <=1.\n\nconstruct f(x)=x-cos(cos(x)), then f'(x)=1-sin(cos(x))*(-sin(x)). \n\nObviously f'(x)>0 when -1<=x<=1.\n\nTherefore f(x) is monotonic increasing at [-1,1].\n\nf(0)=0-cos(cos(0)) = -cos1 < 0;\nf(1)=1-cos(cos(1)) > 0;\n\nSince f(x) is monotonic increasing, then must be only one root between [0, 1].\n\nIt is also obvious that if cos(x) = x, then cos(cos(x)) = x; \n\nTherefore, cos(x)=x is the only root for f(x).
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard