by which method u can integrate this \n((x^2)sinx/(1+x^6)) from (-90)to(90)???\nplz show all steps
218279
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
431721
I believe Integration by Parts would be your best bet here, since (a) it's straightforward to differentiate Arcsin(x), and (b) it's not hard to integrate x^(-2).\n\nLet u = Arcsin(x) and let dv = x^(-2) dx\nThus du = 1/sqrt(1-x^2) dx and v = -1/x\n\nSet up the IBP formula:\n\n Int [Arcsin(x)/x^2 dx] = uv - Int [v du]\n  =  Arcsin(x)/x - Int [-1/(x sqrt(1-x^2)) dx]\n  = Arcsin(x)/x + Int [ x^(-1) (1-x^2)^(-1/2) dx]\n... (by substitution*)\n  =  Arcsin(x)/x + ln(x) - ln (sqrt (1-x^2) + 1)\n\nBe SURE to visit the Integrator at Wolfram.com (see link below)!\n\n* See also worked-out example #6 at:\nhttp://mathforum.org/mathtools/cell/c,15.14.6,ALL,ALL/
64022
Assume n>2 otherwise problem is simple\nI will use "Int" for integral sign\n\nA(n)= Int (tan ^n x) dx = Int (tan^ (n-2) x tan^2 x dx)\n\ntan^2 x = sec^2 x -1\nA(n) = Int tan^(n-2) x sec^2 x dx - Int tan^(n-2) x dx\n\nA(n) = A1(n) - A(n-2)\n\nA1(n) = Int tan^(n-2) x sec^2 x dx\nPut tan(x)=t\nsec^2 x dx = dt\n\nA1(n) = Int t^(n-2) dt = t^(n-1)/(n-1) \n= (tan x)^(n-1)/(n-1)\nThus we have the relation\n\nA(n) = (tan x)^(n-1)/(n-1) - A(n-2)\n\nEvaluate \n\nA(1) = Int tan x dx = -ln cos x +c\nA(2) = Int tan^2 x dx = Int (sec^2 x - 1)dx = tan x -x\n\nNow continue with the rucursion given above for n>2
99151
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
184783
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
451487
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
190687
If you took the 3 from the denominator, you'd have\n\n1/3 int{(x^2 + 2x - 2) / (sqrt(x)}\n\nsince you can move constants in and out of an integral with no problems.  If you got rid of the radical, you'd then have\n\n=1/3 int{(x^2 + 2x - 2) / x^(1/2)}\n=1/3 int{x^2/x^(1/2) + 2x/x^(1/2) - 2/ x^(1/2)}\n=1/3 int{x^(3/2) + 2x^(1/2) - 2x^(-1/2)}\n=1/3[(2/5)x^(5/2) + 2(2/3)x^(3/2) - 2(2)x^(1/2)]\n=(2/15)x^(5/2) + (4/9)x^(3/2) - (4/3)x^(1/2)\n\nEvaluating from 1 to 8 gives you\n\n=[(2/15) x (8)^(5/2) + (4/9) x (8)^(3/2) - (4/3) x (8)^(1/2)]\n-[(2/15) x (1)^(5/2) + (4/9) x (1)^(3/2) - (4/3) x (1)^(1/2)]\n\n= [256sqrt(2)/15 + 64sqrt(2)/9 - 8sqrt(2)/3] - [(2/15) + (4/9) - (4/3)]\n\n= 968sqrt(2)/45 - 34/45\n\n= (968sqrt(2)-34) / 45
532321
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
247390
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.
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.
122982
By sin(x)^5, I assume you mean (sinx)^5.  \nYou mean which of the following?\nFind sinx and then raise it to the fifth power or\nraise x to the fifth power and then take the sin of that number.\nI will work this out the first way as I do not know offhand how to do the second one.\nint = 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\n\nPlease note that the following is NOT true.\nIntegral [ sin(x)^5 ] = -Cos(x)^5 + C\nIf you take the derivative of -Cos(x)^5 (you have to use the chain rule) you will see why.
374338
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
196518
Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
433307
When I have problems with an indefinite integral I use some of my symbolic software to get straight to answer or a "Table of Integrals".  And then I focus on how to get to that particular answer, whether it is through substitution, changing it to a double integral in polar coordinates, etc.\n\nA "free" online integrator program that is offered by Wolfram (maker of Mathematica) is located at\n\nhttp://integrals.wolfram.com/\n\nType your particular problem as\n \nArcTan[1/x] \n\nand it will give you an answer that is solved symbolically for an "indefinite" integral which is\n\nx(arctan(1/x)) + (1/2)(log((x^2) + 1))\n\nHowever to be honest with you it was a few years back when I saw this particular integral, and I honestly don't remember the process.  I think it was a definite integral though.
176771
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
404570
Break up the function into two pieces.\nOne piece from -∞ to 0\nand the other piece from 0 to ∞\n\nIn the first piece, when x is negative,  |x| = -x\nin the second piece, when x is positive,  |x| = x\n\nSubstitute these values in for |x| in each piece and integrate each separately, evaluate, then add the results together.\n\nIn the first piece, the function becomes exp(tx - (-x)) \n= exp(x(t+1))\n\nIn the second piece, the function becomes exp(tx - x)\n= exp(x(t-1))\n\nIt looks like the definite integral diverges however...\n.
657185
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)