who can solve the integration of    sin(x)/x   which is an undetermined integral?
105048
Directly, there is no solution.  If you use a series expansion around x=0, \nsin(x)=sum((-1)^n(x^(2n+1))/(factorial(2n+1)))\nDivide this series by x, then integrate.  After the 3rd term, the error term goes to the ten millionths digit, so I'd say you're pretty safe :)\n\nsin(x)/x = 1 - (x^2)/6 + (x^4)/120 - (x^6)/5040 + .....\nIntegrated is:  x - (x^3)/18 + (x^5)/600 - (x^7)/32580 +\nor \nsum((-1)^n)(x^(2n+1))/(2n+1)*factorial(2n+1))
785756
well sin x = x - x^3/3! + x*5/5! - x^7/7! + .... + (-1)^n x^(2n-1)/(2n-1)! + ....\n= ∑(-1)^n x^(2n-1)/(2n-1)!\n\nSo x sinx = ∑(-1)^n x^(2n)/(2n-1)! (= x^2 - x^4/3! + x*6/5! - x^8/7! + ... )
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
860131
It's undefined at the endpoint of the interval (pi/2).\nFirst go ahead and integrate it like you normally would.\nintegral of sin(x)/sqrt(cosx) dx\nUse substitution.  Let u = cos x, then du = - sinx dx\n= integral of -du/ sqrt(u) \n= integral -u^(-1/2) du\n= -u^(-1/2 +1) / (-1/2 + 1) \n= -u^(1/2) / (1/2)\n= -2u^(1/2)\n= -2 sqrt(u)\nReplace u with cosx\n= -2 sqrt(cos x)\nNow plug in pi/2\n- 2 sqrt(cos pi/2)\n= -sqrt(0)\n= 0\nPlug in pi/4\n-2 sqrt(cos pi/4)\n= -2sqrt( sqrt(2) / 2)\nNow subtract them\n0 - (-2sqrt(sqrt(2) /2))\n= 2sqrt( sqrt(2) / 2)
202670
There is no closed form of this integral. Expressions of this form are power towers, and you can use series methods to evaluate this integral. The value of this integral can be found in the source link. :)\n(To a poster above: All integrals are real, it's just that some may not be expressible in closed form.)
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
848587
sin^2x=(1-cos2x)/2\nantiderivative of (1-cos2x)/2\n=1/2x-(1/4)sin2x +c
706536
Integral: ∫ [2/{(e^2x)+4}] dx\n\nFirst consider the expression: 2/{(e^2x) + 4}\n\n= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]\n\n= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]\n\n\nHence, ∫[2/{(e^2x) + 4}] dx\n\n=  ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx\n\n=  ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx\n\n=  (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx\n\nLet {e^(2x) + 4} = z. Then dz = 2e^(2x) dx\n\nHence the given integral becomes:\n(1/2) x - (1/4)∫(dz/z)\n\n= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration\n\nNow, substitute the value of 'z' in terms of 'x', thus the given integral becomes\n= (1/2) x - (1/4)ln|e^(2x) + 4| + c
696867
cos[x]dx=dsin[x] therefore the integral is over e^y in the limits y=sin[Pi/2]=1 to y=Sin[0] = 0; and is equal to 1-Exp[1]
719088
Substitute u=pi/2-x. Then tan(x)=cot(u), du=-dx and the limits of integration are reversed. So, if we let I_n be the value of the integral, we also have\nI_n is the integral of 1/(1+cot^n (u))du=\nintegral of tan^n(u)/(1+tan^n (u)) du.\n\nNow replace u by x and add to the original to get that\n2I_n =int (1+tan^n (x))/(1+tan^n(x)) dx=pi/2.\n\nThus I_n=pi/4.
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
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
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards
631241
the integration by part is very long and complicated\ni used it many times\nlet u =logsinx          dv=dx\n   du= cosx/sinx          =x\nxlogsinx-int{xcosx/sinx}\nint{xcosx/sinx}  by sbutitution\nlet u=sinx   x=sin^-1(x)\ndu=cosx dx dx=du/cosx\nyou will get int{sin^-1(u)/u} remmeber u=sinx\nby part\nlet k=1/u          dv=sin^-1(u)\n     -1/u^2        int{sin^-1(u)}by part\n\nint{sin^-1(u)}\nlet m=sin^-1(u)                dv=du\n     =1/sqt(1-u^2)                u du\n=usinx^-1(u)-int{u/spt(1-u^2)}\nint{u/spt(1-u^2)} int this by subitution\n\nthen u will get\nsinxsin^-1(x)+(1/2)(1-sin^2x)\n\nso the final ans\nxlogsinx-sinxsin^-1(x)\n+(1/2)(1-sin^2x)+c\n\nplz try to answer this by yourself ,i hope i gave you the idea,thankx
839872
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
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.
888297
You can of course use a brute force approach to get the derivative but a little finesse often can simplify things.  You have:\n\ny=(sin(x)/1-cos(x))^2 = (sin(x)^2)/(1-cos(x))^2\n\nSince sin(x)^2 = 1 - cos(x)^2 it follows that:\n\ny = (1 - cos(x)^2)/(1 - cos(x))^2\n\nBy doing a little algebra up fornt, the final result is quite a bit simpler than some of the others given.\nFactoring the numerator:\n\ny = (1 - cos(x))*(1 + cos(x))/(1- cos(x))^2 = (1 + cos(x))/(1 - cos(x)) \n\nNow your function is of the form\n\ny = (1 + u)/(1 - u) where u = cos(x)\n\ndy/dx = dy/du * du/dx = (1/(1 - u) + (1 + u)/(1 - u)^2)(-sin(x))\n\ndy/dx = -2*sin(x)/(1 - u)^2 = -2*sin(x)/(1 - cos(x))^2\n\nBy doing a little algebra up front, the final result is quite a bit sinpler than those above.