Does anyone know the intergral of (Sinx)^5 dx ?
414817
int = integral\nint[(sinx)^5dx]\nRewrite (sinx)^5 as sinx (sinx)^4\n(sinx)^4 = ((sinx)^2)^2\n(sinx)^2 = 1 - (cosx)^2\nThus, \n(sinx)^4 = ((sinx)^2)^2\n= (1-(cosx)^2)^2\nNow the integral is\nint[sinx(1-(cosx)^2)dx]\nUse substitution.\nLet u = cosx, then du = -sinxdx\nNow the integral is\nint[-(1-u^2)^2du] \nExpand (1-u^2)^2 and multiply through by the -1\n-(1-u^2)^2 = -(1 - 2u^2 + u^4) = -1 + 2u^2 - u^4 \nThe integral now equals\nint[(-1 + 2u^2 - u^4 )du]\n= -u +(2/3)u^3 - (1/5)u^5 + C, where C is a constant\nReplace u with cosx to get the final answer.\n-(cosx) +(2/3)(cosx)^3 - (1/5)(cosx)^5 + C
160524
This is not a calculus problem, this is merely a trigonometric identity.  No derivatives, integrals, or numerical approximation were used.\n\nDefinition:\n\ntanh(2x) = sinh(2x)/cosh(2x)\n\nYou know the double angle formulas for sinh and cosh:\n\nsinh(x + x) = sinh(x)cosh(x)+cosh(x)sinh(x) = 2cosh(x)sin(h)\n\ncosh(x + x) = cosh(x)cosh(x)+sinh(x)sinh(x) = cosh(x)^2 + sinh(x)^2\n\nDivide:\n\n(2cosh(x)sinh(x))/(cosh(x)^2 + sinh(x)^2)\n\nDivide numerator and denominator by cosh(x)^2:\n\n((2cosh(x)sinh(x))/cosh(x)^2) / ((cosh(x)^2 + sinh(x)^2)/cosh(x)^2)\n\nSimplify:\n\n(2 sinh(x)/cosh(x))/(1 + sinh(x)/cosh(x)) \n\nSimplify more to solution:\n\n2tanh(x)/(1 + tanh(x)
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
816208
RHS  (simplify it into sin and cos terms)\ncsc^2(x)/cotx\nsinx/cosx*1/sin^2x\nsinx/sinx*1/cosxsinx\n2/2sinxcosx\n2/sin2x\n2csc2x\nhence LHS=RHS
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C
997471
IF you want to try and get it in that EXACT form dicatated by Mathematica, I suggest you start with u=1+sin(x) and then solve for x in terms of u and do some resubstitution.  You'll get the arcsin function, which becomes a polynomial in terms of another dummy variable when you start the first u*dv substitution.  Notice I said "FIRST", because to get it in the form that you want, you'll have to do it about 6-10 times.  Also, you'll have to do some "right triangle simplification" throughout the problem as well.  Of course, then you get to factor it.  Have at it.  I did this very thing, stopping at the 3rd u*dv substitution because it is just ridiculous.  \nI did, however, get several of the terms.  I think if you decide to do this, you're going to need lots of paper.  I was doing it symbollically in a CAD program to keep things organized and such, and it was already 5 pages.  Oh, and not to mention that this "answer" given by Mathematica isn't even a closed form "answer".  It left it in terms of a polylogarithm, which is still an integral hidden in notation.  \n\nThe point is that the fuction (among others) is best integrated numerically because there is no useful closed form for this function.  This is what I did and it took me 2 minutes.  \n\nx^2/(1+sin(x))=x^2 - x^3 + x^4 - (5/6)x^5 + (2/3)x^6 - (61/120)x^7 + (17/45)x^8 - (277/1008)x^9 .................\n\nIntegrate termwise, which is fricking easy.  \n\n(1/3)x^3 - (1/4)x^4 + (1/5)x^5 - (5/36)x^6 + (2/21)x^7......\n\nNotice it is not undefined (like the original function) and the error term goes to 10^-7 after the first 5 terms.  I'd say that's as accurate as anyone would ever really need to be, and if not, they can always expand this function further.\n\nTell your uncle he's a bastard just for me.  He's teasing you.
227193
∫[(2x^4) - (4/x^5)]dx=\n\n∫(2x^4)dx - ∫(4/x^5)dx=\n\n2∫x^4dx - 4∫[x^(-5)]dx=\n\n2*[x^5/5] - 4*[x^-4/-4]=\n\n(2/5)x^5 + x^-4 + C=\n\n(2/5)x^5 + 1/x^4 + C
1079454
The previous answers are all WRONG. It is not easy to give a general formula, but here is the definition and the proceedure:\n\nFirst of all, remember that the power series expansion of sin x is\nsin x=x -(1/6) x^3 +(1/120)x^5-+...\n\nIf you plug in At for x, you get\nsin(At)=At-(t^3/6)A^3 +(t^5/120)A^5-+...\n\nThe powers here are obtained by matrix multiplication. For a *diagonal* matrix, you can simply take the sine of the diagonal entries (as the previous posters suggested), but that doesn't work for non-diagonal matrices. It should be pointed out that exp(At) or cos(At) would be defined by power series in a similar way.\n\nIn most applications, the matrix A is symmetric, so can be diagonalized (by finding eigenvalues, etc). In this case, there is\na diagonal matric D and another matrix P so that\nA=PDP^{-1}.\nThe matrix D has the eigenvalues of A along the diagonal and the matrix P has the corresponding eigenvectors down the columns.\nAnyway, for such matrices,\nsin(At)=P{sin(Dt)]P^{-1}.\nSince sin(Dt) is easy, this allows the computation of sin(At).\n\nFor non-diagonalizable matrices, the computation is harder.
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
571352
I am assuming I means intergral, the way you find the areaunder a curve.\n\nA = Integral of (x^3+4x-3.6) = x^4/4 + 2x^2-3.6x\n\nusing the formula Integral of (x^n) = x^(n-1)/n\n\nOf course you need limits of integration. i.e. between x=2 & x=5 to get a numerical answer. i.e. 23.5\n\n\nUse these already proven exponent rules.\n\na^0 = 1,    so 5^0 = 1 \n\na^-n = 1/a^n,    so 5^-1 = 1/5^1 = 1/5 or 0.2\n\n\nAlso:\n5^0 = 1 because of the following exponent rule & division\n\na^m/a^n = a^(m-n) \n\nso 5^3/5^3 = 5^(3-3)= 5^0 \n\nhowever  \n\nso 5^3/5^3 = 125/125 = 1\n\nTherefore\n\n5^0 = 1\n\nThe exception to the rule is 0^0, because you can't divide by zero, although wriiten as a function its limit approcahes 0. But that's another class.\n\nYou can use the same process to come to the answer for\n\n5^-1 \n\n5^2/5^3 = 5^-1 = 25/125 = 1/5 or 0.2\n\nThese rules however are already proven.\n\na^0 = 1\n\na^-n = 1/a^n
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
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))
978070
Try\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm\n\nYou can also try http://www.helpalgebra.com\n\nGoodLuck!
324244
sinx + xcosx  (product rule)
969937
'\nu= sec x \ndu = tan x sec x dx\ntan x dx = du/sec x = du/u\ntherefore:\nintegral of tan x dx  = integral of du/u = ln |u| + c = ln| sec x| + c
871386
∫udv = uv - ∫vdu\n∫x^2dx/√(1-x)\nlet u = x^2, dv = dx/√(1-x)\ndu = 2xdx, v = -2√(1-x)\n∫x^2dx/√(1-x) = -2x^2√(1-x) + ∫(4x√(1-x))dx\nu = 4x, dv = √(1-x))dx\ndu = 4dx, v = -(2/3)(1-x)^(3/2)\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) + ∫(8/3)(1-x)^(3/2)dx\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) - (16/15)(1-x)^(5/2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - (8x/3)(1 - x) - (16/15)(1-x)^2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - 8x/3 - 8x^2/3 - (16/15)(1 - 2x + x^2)\n∫x^2dx/√(1-x) = (1/15)√(1-x)(30x^2 - 40x - 40x^2 - 16 + 32x + 16x^2)\n∫x^2dx/√(1-x) = -2√(1 - x)(x(3x + 4) + 8)/15