How to integrate cos (x^2-9) with respect to x.
210425
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.
473009
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
365858
sec(-x)cos(-x); cos(-x)=cos(x); sec(-x)=sec(x)\nsec(x)cos(x); sec(x)=1/cos(x)\n[1/cos(x)]cos(x)\ncos(x)/cos(x)=1
448287
You need to integrate that formula. First of all, expand it:\n(3-√x)²/√x\n(9 - 6√x + x)/√x\n9/√x - 6 + x/√x\n9x^-0.5 - 6 + x^0.5\nNow integrate:\ny = 18x^0.5 - 6x + 2/3 x^1.5 + c\nSubstitute in the point: \n2/3 = 18 - 6 + 2/3 + c\nc = -12\nSo y = 18x^0.5 - 6x + 2/3 x^1.5 - 12.
479974
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
2514
∫xSec(x²)²dx = .5Sin(x²)/Cos(x²)
157237
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...
359458
Int tanx ln(cosx)dx\n\nwrite tanx=sinx/cosx\n\nsubstitute y = cos(x)\ndy=-sinx dx\n\n-Int ln(y)/y dy\n\nsub z = ln(y)\ndz=ln(y)/y dy\n\n-Int z dz = -z^2/2 = -(1/2)ln^2(y) = -(1/2)ln^2(cos(x)) + C
397740
Set u = Arccos(x) such that dx = sinu du.  Then the integral becomes the integral of (u^2)sinu du.  Now the integral of this is (by parts) = -(u^2)sinu + 2*(the integral of) (usinu du).  And the integral of usinu (by parts) = -ucosu+(the integral of)(cosu du).\n\nHope this helps,\n\nSteve
3687
e^(ix)=cos(x)+i*sin(x)\n\nx^i=e(i*ln(x))\n\nx^i=cos(ln(x))+i*sin(ln(x))
464530
∫x³/√(x² + 9)dx Let x = 3 tanθ therefore dx = 3 sec²θ . dθ\n=∫27 tan³θ/√(9 tan²θ + 9) . 3 sec²θdθ\n= 27 tan³θ/√(tan²θ + 1) . sec²θ . dθ (tan²θ + 1 = sec²θ)\n= 27∫tan³θ/secθ . sec²θ . dθ \n= 27∫tan³θsecθ . dθ \n\n= 27∫tanθsecθ (sec²θ + 1) . dθ \n= 27∫(sec²θ. tanθsecθ + tanθsecθ) . dθ (NOTE dsecθ/dθ = tanθsecθ)\n= 27( ⅓sec³θ + secθ) + c\nIf x = 3 tanθ then tanθ = x/3 so cosθ = 3/√(x² + 9)\nhence secθ = √(x² + 9)/3\n\nThus required integral = ⅓(x² + 9)^(3/2) + 9(x² + 9)^½ + c
482362
Integrate from -π/2 to π/2 and double it
15464
When you are doing integrations, all you are doing is computing an area under a curve. Sometimes the formula for the area will be obvious to you, from something you already know or recognize. Othertimes, you have to transform the problem into one you know already. \n\nThis is the method called integration by substitution. \n\nFor example, suppose you want to integrate the function sqrt(1 - x^2) with respect to x, between some limits. Most times, you will not remember or recognize this integral. However, you might remember that sqrt(1-(sin a)^2) is cos a, which you know how to integrate, and therefore if you let x=sin a, you have transformed the integral into one you recognize and can easily solve. \n\nThere is a HUGE amount of material written on this on the internet and in books, but this should get you started. All you are doing is transforming the problem into one that you can solve easier.
367997
the integral of [cos 2x dx] is [1/2 sin 2x + c]
445320
sure. no problem, pal.\nx^2 + 9x + 20
92095
3cosx-cosx-2=0\n2cosx=2\ncosx=1\nx=0 or x=2π (on domain from o to  2π)
207416
Go to http://mathworld.wolfram.com/Cosine.html