what is the result of indifinte integral (sin(x)/x)dx? are there some functions we cant find infinte integral?
339966
The integral of sin(x)/x cannot be written in finite terms with the 'elementary functions', which consist of trig, inverse trig, exponential, logarithms, and algebraic functions. It can be written as a power series or in other ways, depending on what you want to do with it. Good amth programs will compute values for this integral (as approximations).\n\nAs for the integral of sin(x)/x from 0 to infinity, the result is pi/2, but that usually is done with complex analysis. It cannot be done easily with elementary methods.
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
847056
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
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
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
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
45131
Calculus was confusing enough in English - I can't imagine taking it in Japanese.  You might want to try to get hold of a textbook or something in English.\n\nIt's an integral.  More specifically an indefinite integral.  It's kind of like the inverse of the derivative function except for that there is no way to get the constant back (hence the "C").  Basically the indefinite integral is the set of all functions whose derivative is equal to the thing between the integral sign and the dx (or dy, dq, whatever the variable is... I'm blanking on general name of the thing).  The integral symbol tells you it's an integral (obviously), and the "x" in dx tells you what variable you're integrating with respect to (that will make more sense when you learn to integrate more complicated expressions).  So the indefinite integral of x^2 is 1/3x^3 + C with respect to x where C is some constant because the derivative of 1/3x^3 + C is x^2 (remember that the derivative of a constant is zero).\n\n***A Note on Definite and Indefinite Integrals***\nA couple of the answers below appear to be describing the definite integral.  The definite integral can be described as the sum of the areas of rectangles, and can be used to find the area under a curve (among other things).  The indefinite integral (or antiderivative), however, which is the one refered to in the question, describes a function.  Although the function described by the indefinite integral is extremely useful in calculating the area under a curve, it does not directly describe this area.  In addition to being used to caculate the definite integral, the indifinite integral can also be used, for example, to find an equation for the position of an object given an equation for its velocity.\n\nAnother afterthough: I may have gotten some of the terminology wrong as my high school math department had a strange affinity for inventing their own terms.
212677
Little reply: No.\n\nP.S. 12.08 AM GMT\nThis gives the Fresnel Integral S(x), more exactly\n\nS(x)=Integral(.5Pi*x^2)\n\n(the integral is from 0 to x), and it is not possible to reduce it in a finite way.\n\nHowever S(x) has an asymptotic expansion for x>>1\n\n.5-(cos(.5Pix^2))/(x*Pi)\n\nwhich gives limit S(infinity)=0.5
385662
If you mean to use the Riemann integral, the function is not integrable since it is not bounded. For the Lebesgue integral, the rationals are a set of measure 0, so are irrelevant. In the interval (1/(n+1), 1/n], the function has value n, so the integral over that interval is [1/n-1/(n+1)]*n=1/(n+1).\nThe sun of 1/(n+1) from 1 to infinity diverges to infinity.
959491
sin(X)+sin(2x)+sin(3x)+....+sin(nx)\n= img[exp(ix)+exp(i2x)+exp(i3x)+...+exp(inx)]\n=img[(exp(i(n+1)x) -1) / (exp(ix) - 1)]\n=img[(exp(i(n+1)(x/2)) -1) /exp(i(n+1)(x/2)) * (exp(i(n+1)(x/2)) -exp(-i(n+1)(x/2))) / (exp(i(x/2)) - exp(-i(x/2)))\n=[sin((1/2)(n+1)x)/sin((1/2)x)]* img(exp(inx/2))\n=[sin((1/2)nx)/sin((1/2)x)]*[sin((1/2)(n+1)x)/sin((1/2)x)]/[sin((1/2)x]\n\n[exp is exponential, img is imaginary part, 'i' is usual imaginary i in complex numbers] I also suggest you write down this so that you will be more clear with the equations. Sorry but without equation editorits bit difficult to type math.\n\nIf you want an infinite sum, please take the limit as x-->infinity of the last term obtained.\n\nI hope this answeres your question. All the sest.
149854
babymegan101 is wrong because the derivative of xsin(ax^2+b)+C is sin(ax^2+b) + 2ax^2*cos(ax^2+b). (And we expect the derivative to be sin(ax^2 + b)). I've tried every trick in a physicist's toolbox to solve this integral but I have not succeeded. Therefore I propose that there isn't a definitive answer to the question, and that it can only be solved by approximation using numerical methods.
700700
try putting x^2 = i*t\nwhere i = complex iota\n t is varible \n\nthere integral will be in form : i/2*( cos(t)+i*sin(t) ) dt\n\nso answer will be : 1/2 * (i* sin (t) + i*cos(t) ) = 1/2 * e^ (i*t) = 1/2 * e^x^2\n\nThat's the answer....... :D\n\n\n[Assuming u know that e^(i*t) = cos(t)+i*sin(t) ]\n\ncan be solved easily ......... try it
950010
replace [Cos(2x)]^2 with 1 - [Sin(2x)]^2.\n\nnow multiply to get...[Sin(2x)]^3 - [Sin(2x)]^5.\n\nthen separate integral into two parts...\n\nnow use power rule and chain rule and don't forget to add a constant.
761317
We have to find derivative of, y = the integral where lower limit is e^x and upper limit is 0 of sin^3 t dt.\n\ny = int(e^x)(0) [sin^3 t] dt     [I would write lower limit in first bracket and upper limit in bracket after that, so dont confuse]\n\nNow, the derivative of the integrand f(t)=sin^3 t is f'(t)=3 (sin^2 t) *cos t\ndy /dt= d {int(e^x)(0) [sin^3 t] dt }/dt\n=>dy /dt= d {f(t)}/dt (at 0) - d {f(t)}/dt (at e^x)\n=>dy /dt= f'(t) (at 0) - f(t) (at e^x)\n=>dy /dt= [3 (sin^2 t) *cos t] (at 0) - [3 (sin^2 t) *cos t ] (at e^x)\n=>dy /dt= [3 (sin^2 0) *cos 0] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= [3 (0) *1] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= -[3 (sin^2 (e^x)) *cos (e^x) ]\n\nI hope that answers your question.\n\nAll the bset
377430
Differentiating sinx gives cos x.\n\nDifferentiating sin[f(x)] gives f'(x)cos[f(x)]\n(The chain rule)\n\nf(x) = 5x\n\nso f'(x) = 5\n\nHence:\n\n5cos(5x)
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.