integration of """e^sinx.dx"""?
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.
171290
I think you mean "integral" of x^(n-1)* exp(-(log)^2)/2 and I THINK that the e and the log should cancel so you would have X^(n-1) * -X^2/2 = x^n-1+2 /2 so the integral of that would be... X^(n+1)/(2n+4)
989233
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out.
769357
It can be helpful here to express this in terms of complex exponentiation. Note that sin x = (e^(ix) - e^(-ix))/(2i) and cos x = (e^(ix) + e^(-ix))/2. So:\n\n(sin x)^3 * (cos x)^4\n(e^(ix) - e^(-ix))^3/(2i)^3 * (e^(ix) + e^(-ix))^4/2^4\n(e^(3ix) - 3e^(ix) + 3e^(-ix) - e^(-3ix))/(-8i) * (e^(4ix) + 4e^(2ix) + 6 + 4e^(-2ix) + e^(-4ix))/16\n-1/(128i) * (e^(7ix) - e^(-7ix) + e^(5ix) - e^(-5ix) - 3e^(3ix) + 3e^(-3ix) - 3e^(ix) + 3e^(-ix))\n-1/64 (sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x))\n\nThus, all that tedious mucking about in the complex numbers has allowed us to reduce a hard-to-integrate product of powers of trigonometric functions into a sum of very to simple integrate functions. Thus:\n\n∫sin^3 x cos^4x dx\n-1/64∫sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x) dx\n-1/64 (- cos (7x)/7 - cos (5x)/5 + cos (3x) + 3 cos x) + C\ncos (7x)/448 + cos (5x)/320 - cos (3x)/64 - 3 cos x/64 + C
807024
You need to use integration by parts for this one as well.\nLet's call the integral A; that is, \nA = integral [(e^-x)cos(2x) dx]\nLet u = cos(2x) and dv = e^-x dx, then\ndu = -2 sin(2x) and v = -e^-x\nA = uv - integral [v du]\nA = -(e^-x)cos(2x) - integral [2(e^-x)sin(2x) dx]\nNow we need to use integration by parts again.\nLet u = 2sin(2x) and dv = e^-x dx, then\ndu = 4cos(2x) and v = -e^-x dx\nuv - integral(v du)\n= -2(e^-x)sin(2x) + integral [4(e^-x)cos(2x) dx]\n= -2(e^-x)sin(2x) + 4*integral [(e^-x)cos(2x) dx]\nSo we have that \nA = -(e^-x)cos(2x) - integral [2(e^-x)sin(2x) dx]\nA = -(e^-x)cos(2x) - [-2(e^-x)sin(2x) + 4*integral [(e^-x)cos(2x) dx]]\nA = -(e^-x)cos(2x) + 2(e^-x)sin(2x) - 4*integral [(e^-x)cos(2x) dx]]\nNotice that the last integral is actually the integral we started with.\nA = -(e^-x)cos(2x) + 2(e^-x)sin(2x) - 4A\nSolve for A by adding 4A to each side\n5A = -(e^-x)cos(2x) + 2(e^-x)sin(2x)\nMulitply both sides by (1/5)\nA = (1/5)[-(e^-x)cos(2x) + 2(e^-x)sin(2x)]\nA = (1/5)(e^-x) [2sin(2x) - cos(2x)]
980558
∫xe^x dx \n= ∫x d(e^x)/dx dx \n= xe^x - ∫e^x (dx/dx) dx\n= xe^x - ∫e^x dx\n= xe^x - e^x + c\n= (x - 1)e^x + c
1051703
collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
793695
' the integral of 1/e^u du = integral of e^{-u}du\n=-e^{-u} + C
158690
substitute x=e^y\nthen use a table of integration formulas?
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
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
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
861008
NO\n\nf'(x)=(-sinx)(e^cosx)\nf'(4.0)=(-.07)(2.7)= -1.89
761146
You're trying to equate them.  What you really need to do is take their difference and then take the integral of that term:\n\nA = ∫(lnx - (x-2))dx = ∫lnx dx - ∫(x-2)dx\n\nLooks a little easier now, eh?\n\nThere's still a problem.  To find the limits of integration, you have to solve the 2 eqns simultaneously to get the intersection points.  I don't think there is an algebraic way to do that, so here are the limits (determined numerically); x = .158594 and x = 3.1465
159947
What is "CAS"?\n\nAnyway, one way is to know that \n   sin(x) = (e^(ix) - e(-ix))/(2i) \nwhere i is the imaginary number.\n\nSome numerical fidgetry leads to:\n\n   - e^(-2x)(cos(x) + 2sin(x))/5\n\nYou can probably get the same result by integrating by parts twice:\n\n    I = int(sin(x)e^(-2x)dx)\n\nUsing u = sin(x), dv = e^(-2x)dx we get v = -e^(-2x)/2\n\n    I = -sin(x)e^(-2x)/2 + 1/2* int(cos(x)e^(-2x)dx)\n\nNow applying integration by parts on the second term:\n\n    J = int(cos(x)e^(-2x)dx)\n\nwith u = cos(x) and dv = e^(-2x), we get du = -sin(x) dx\nand v = -e^(-2x)/2\n\n   J = -cos(x)e^(-2x)/2 - 1/2 int(sin(x)e^(-2x)dx)\n\nSo:\n   I = -sin(x)e^(-2x)/2 + 1/2 J\n\nand:\n\n  J = -cos(x)e^(-2x)/2 - 1/2 I\n\nSolving for I will give you the same result.  [ The integration by parts steps might not be exactly correct - check them yourself.]
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.
785428
y=e^cos(x) + cos(e^x)\ny'=-sin(x)e^cos(x) + -sin(e^x)e^x\n=-(sin(x)e^cos(x)+sin(e^x)e^x)