How to integrate cos (x^2-9) with respect to x.?
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.
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
564945
4*sin(x)cos^2(x) \n= 2*{2*sin(x)cos(x)}cos(x) \n= 2*sin(2x)cos(x)\n= sin(3x) + sin(x)\nTherefore, sin(x)cos^2(x) = [sin(3x) + sin(x)]/4\n\n4*sin^2(x)cos(x)\n= 2*{2*sin(x)cos(x)}sin(x)\n= 2*sin(2x)sin(x)\n= cos(x) - cos(3x)\nTherefore, sin^2(x)cos(x) = [cos(x) - cos(3x)]/4\n\nHence the expression inside the integral sign is\n[sin(3x) + sin(x) - cos(x) + cos(3x)]/4\n\nThus the integral becomes\n(1/4)*∫[sin(3x) + sin(x) - cos(x) + cos(3x)]dx\n= (1/4)*[∫sin(3x)dx + ∫sin(x)dx - ∫cos(x)dx + ∫cos(3x)dx]\n= (1/4)*[{-cos(3x)}/3 - cos(x) - sin(x) + {sin(3x)}/3] + C\nwhere C = arbitrary constant of integration\n= (1/4)*[{sin(3x) - 3*sin(x)} - {cos(3x) + cos(x)}]/3 + C\n= (1/12)*[- 4sin^3(x) - 4cos^3(x)] + C\n[N.B. cos(3x)=4cos^3(x)-3cos(x) and sin(3x)=3sin(x)-4sin^3(x)]\n= - [sin^3(x) + cos^3(x)]/3 + C
640359
let sinx=u\nthen cosxdx=du\n\nand ∫[(cosx)/(9+sin^2x)]dx=\n\n∫du/(9+u^2)=        \n\n\n ( now use this formula\n∫du/(a^2+u^2)= 1/a arctan(u/a)+C)\n\n\n=1/3 arctan(u/3) +C=\n\n\n=(1/3)arctan(sinx/3)+C
10077
e^(ix)=cos(x)+i*sin(x)\n\nx^i=e(i*ln(x))\n\nx^i=cos(ln(x))+i*sin(ln(x))
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
532184
One method is to integrate by parts twice. There is another way where you can use complex methods: \ncos2x = Re e^(2ix)\n[Re means the real part]\n\nIntegral e^(-x)cos2x dx\n= Integral e^(-x) Re e^(2ix) dx\n= Re Integral e^((2i-1)x) dx\n= Re e^((2i-1)x)) / (2i-1) + C\n= Re e^(-x) e^(2ix) / (2i-1) + C\n\nWrite the above in the standard A + iB and then you can find your answer. Good luck.
1154463
cos(x)*cos(x) = cos^2(x)= 1-sin^2(x)\ntherefore the problem is now down to 1-sin(x)-Sin^2(x)\nthen integrate each one separately. \nfor inter of 1 is just x\nintegral of  -sin(x) is cos(x)\nand for sin^2 use the half angle identity \nsin^2(x) = .5*(1-cos(2X))\nand therefore -sin^2(x) = -.5*(1-cos(2X))\nintergral of which is -.5x+.25sin(2x)\n\nand the final answer add them and you'll get \n.5x + cos(x) + .25sin(2x)\nor \n.5x + cos(x) + .5sin(x)*cos(x)\nbecause sin(2x) = 2sin(x)*cos(x)
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
281867
Well I assume you are trying to work out x and I also assume that it's not cos [to the power of] (x^3) but Cos [multiptied by] (x^3)\n\nYou need to integrate to remove the d/dx:\n\nCos integrated is Sin and you then have to divide by the differental of what's in the brackets\n\n=> sin(x^3) / 3x^2
1128475
If you were to integrate f(x) from 0 to pi/2\n\nit wold be  int(f,0,pi/2).
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
895342
x^2-9x-x+9=0\n(x-9)(x-1)=0\nx=9 or x=1
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
757042
I'm assuming you mean:\n\n∫∫x cosy dy dx\n\nwhere the bounds of the outer integral are x=0 to x=1 and the inner integral are y=0 to y=x².\n\nFirst evaluate the inner integral from y=0 to y=x², pretending x is constant.\n\n∫x [∫cosy dy] dx = ∫x [siny] dx = ∫x [sin(x²) - sin(0)] dy = ∫xsin(x²)dx\n\nThen use u substitution, u = x², x = √u, du = 2x dx, so dx = du/2x = 1/(2√u) du:\n\n∫√u sin(u) 1/(2√u) du = 1/2 ∫ sin(u) du = [-1/2 cos u] = [-1/2 cos(x²)], evaluated from x=0 to x=1\n\n-1/2[cos1 - cos 0] = (1 - cos1)/2\n\nNot bad for not having done calc since 1992, huh?
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.