please find the integral of this function int sqrt(sinx)dx?
611207
The only way to do this is by using the reduction formula for int(sin(x)^n), which is valid for n>0.  Unfortuantely, it reduces to another integral for sin(x)^(n-2), for which the reduction formula no longer is valid.  Hence, the only way to do this integral is by using numerical integration techniques.\n\nProbably the best way to do this problem is to convert sin(x)^(1/2) to a taylor series.  \n\nI did this and got:\nx^0.5 - x^2.5/12 + x^4.5/1440 - x^6.5/24192 - 67/29030400*x^8.5 .....\n\nIntegrate:\n\n(2/3)*x^(3/2)-(1/42)*x^(7/2)+(1/7920)*x^(11/2)-(1/181440)*x^(15/2)-(67/275788800)*x^(19/2)......\n\nNeedless to say, after the first 4 terms, any error from a direct integration of sqrt(sin(x) is less than 10^-6, which I would say is pretty damn close.  \n\nThere ya go...
84808
multiply and divide by (1 - sinx)\n\nyou get (1 - sin^2(x))/(cosx(1-sinx)) that is cos^2(x)/(cosx(1-sinx)) that is cosx/(1-sinx)\n\nnow at the denominator you have a function and at the numerator you have the derivative of that function, save for the sign. \nyour integral is \n-ln(|1-sinx|) + c\n\n---- \n\nproof that we're all saying the same thing:\n- ln |1-sinx| = ln (|1 - sinx|^-1) =\n\n= ln |1/(1-sinx)| ---> divide and multiply again for \n(1 + sinx)  ----> = ln |(1 + sinx)/cos^2(x)|\n\n= ln |(secx + tanx)/cosx| = ln|(secx + tanx)secx| = \n= ln |secx + tanx| + ln|secx|\n\nSo you have many ways to write your answer: you can choose the one which you prefer\n- by elegance\n- by minimization of spread error in calculating it\n- by whatever you like\n\nhave a nice day
420145
The antideriative is the same as an integral.  For this, we can use the fact that sin^2(t) + cos^2(t) = 1.  Then,\n\n[int] cos^2(t)/sin(t) dt = [int] (1 - sin^2(t))/sin(t) dt\n= [int] (1/sin(t)) - sin(t) dt\n= [int] csc(t) - sin(t) dt\n= ln(tan(t/2)) + cos(t) + C
979363
You can't do this integral straight-up because it is a special case.  The integral is:\n\nsqrt(pi/2)Fresnel(sqrt(2/pi)x)\n\nBelow are websites with an Integral Calculator and the Fresnel function\n\n-------\n\nHope this helps
66517
Integral of sqrt(6x) + x^2\n= Integral of 6^(1/2).x^(1/2) + x^2 \n= 6^(1/2). (x^3/2) /(3/2) + (x^3)/3 + constant\n= 2/3. sqrt( 6x^3) + 1/3 . x^3
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`
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)
757042
I'm assuming you mean:\n\n∫∫x cosy dy dx\n\nwhere the bounds of the outer integral are x=0 to x=1 and the inner integral are y=0 to y=x².\n\nFirst evaluate the inner integral from y=0 to y=x², pretending x is constant.\n\n∫x [∫cosy dy] dx = ∫x [siny] dx = ∫x [sin(x²) - sin(0)] dy = ∫xsin(x²)dx\n\nThen use u substitution, u = x², x = √u, du = 2x dx, so dx = du/2x = 1/(2√u) du:\n\n∫√u sin(u) 1/(2√u) du = 1/2 ∫ sin(u) du = [-1/2 cos u] = [-1/2 cos(x²)], evaluated from x=0 to x=1\n\n-1/2[cos1 - cos 0] = (1 - cos1)/2\n\nNot bad for not having done calc since 1992, huh?
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]
402001
Watch it ….! \n\ndelta\n\n x =t^2  ---> t= sqrt(x)\ndx = 2t dt .............(not dx = dt /2t )\n-->\ndt = dx/(2 Sqrt(x)\n\nand this Integral can not be done using Analytical Method; Also, the (method of Integration by Part) can not apply in such integral because either of the cos(t) and sqrt(t) have infinitive derivatives.
294550
Graph sin(2x) and sin(x). Once graphed you will see that y=sin(2x)the top function. You will integrate (sin2x-sinx)dx from 0 to pi. you can split the intergrals\nint(sin2x)dx = -1/2cos2x \nminus\nint(sinx)dx = -cosx\n\n-1/2cos2x+cosx evaluated at pi and 0.\n\n(-1/2 cos(2(pi))+ cos(Pi))-(-1/2 cos(0)+cos(0))
744903
1)\n\ntan^2(x) = sec^2(x) - 1\n\nSo the integral of sec^2(x) + 4 = tan(x) + 4x\n\n2)\n\ncsc(x) = 1/sin(x)\n\nthe integral is then of 1+sin(x)/cos^2(x).\nlet u = cos(x) then du = -sinx dx.\nso the integral is x+sec(x).\n\n3)\n\nlet u = cos(2x+1) then du = -(1/2)sin(2x+1) dx.\n\nso the integral is of -(1/2)/u^2 = 2/u = (1/2)/cos(2x+1) = sec(2x+1)/2\n\nSteve
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
958105
Hi,\n\nThis not just ...sqrt(pi) as one respondent thought.\n\nWhat you need are two very useful tools for integration, part one is solved by changing variables and part two is solved with integration by parts.\n\n1. let x= (y-a)/sqrt(b), so dy/dx = sqrt(b) \n=> dx = (1/sqrt(b)) dy (this isn't strictly correct but it works here)\n\nso:\nI = int(exp(-(x-a)^2/b)dx,-inf..+inf) \nI = int(exp(-((x-a)/sqrt(b))^2)dx,-inf..+inf) \nI = int(exp(-y^2)*(1/sqrt(b))dy,-inf..+inf)\nI=(1/sqrt(b))*int(exp(-y^2)dy,-inf..+inf)\nI=(1/sqrt(b))*sqrt(pi)  (from our given result)\nI = sqrt(pi/b)\n\nThe seond part requries a decent understanding of integration by parts, and if you email me I can go through this with you.
304641
Integral f(x)dx=-3cos(pi)+3cos(0)=3\nSo average value of the func f(x) = 3x sin (x^2) is \n3/sqrt(pi)
832686
Take the integral of this function, like this:\n\n∫ (10 + x²) / x³ =\n\n∫ (10 / x³) + ∫ (x² / x³) =\n\n10 ∫ (1 / x³) + ∫ (1 / x) =\n\n10 (-1 / 2x²) + ln(x) + C = \n\n-5 / x² + ln(x) + C\n\n~ ♥ ~
83265
That's an elliptic integral.  Check out the link.