evaluate the integral of 1/(25-4x^2)^1/2  dx?
347908
I got the same answer as math girl, but did it SLIGHTLY different.  I recognized that the form of the integral was one of the sine inverse functions.  That is, the derivative of Sin^-1 (x) = 1/(1-x^2)^.5  \n\nIf you recognize this relationship, the problem is an easy u substitution away.\nNamely, factor out 25 from the denominator in the square root, leaving you with \ndx/(25(1-(4x^2)/25))^.5\n=dx/(5(1-((2/5)x)^2)^.5)\nThen make a the u substitution of u = (2/5)x, giving dx = (5/2)du\n\nSubstituting back gives (1/5)(5/2) outside of the integral and inside the integral du/(1-u^2)^.5\n\nThe integral is (1/2)Sin^-1 (u) + C\nSubstitute back and get (1/2) Sin^-1 ((2/5)x) + C\n\nWriting it down with symbols on paper makes this method very easy, even if following it in this medium difficult. \n\nThe key being knowing your inverse trig derivatives.  If you don't know them, you are better off using math girl's method.
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C
916838
(1/2)(1/2)=1/4\n\nB. 1/4
753779
When you have written it out twice you can then multiply out the brackets\ni.e. x^2(x^2-1/x^2)-(1/x^2)(x^2-1/x^2)\nthis will give you x^4-2+1/x^4\nRewrite using a negative indice to replace the fraction\nx^4-2+x^(-4) and then integrate as per norm using the rule\nthat the integral of x^n = x^(n+1)/(n+1).\nHope your day improves.
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
796927
try the link below...... hope it can put an end to ur integral problems
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)
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
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...
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
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
737484
5^-2 = (1/5)^2 = (1/5)*(1/5) = (1*1)/(5*5) = 1/25
403275
Look, to count this integral, you must make a subsitution:\n\nu = 1+5t\ndu = 5 dt\ndt = du/5\n\nint 8/(1+5t) dt =\n= int 8/u du/5 =\n= 8/5 * int (1/u) du =\n= 8/5 * ln u + C =\n= 8/5 * ln(1+5t) + C
297440
Are you looking for indefinite  integral   of x^(sin(1/x)) ?\n\nMathematica has a web site you can plug-in the integrant and get the result (if it exist in closed form)\n\nBy hand: try to break this power expression using log and exp functions. If  a=ln(b), then b=exp(a) and therefore b=exp(ln(b)).\n\nIn this way your integrant could be written as \n\nexp(ln(integrand))\nand since it is of power form, use the properties of ln and exp functions to break it down and thus reduce it to simpler integrals.
727341
first integrate\n\nF(x) = x^3 -x^2  from [1, 3]\n= 3^3 -3^2 -1^3 + 1^2\n= 27 - 9 -1 + 1\n= 18\n\naverage = 18/(3-1) \n= 18/2\n= 9
396729
To solve this you need a sum function\nand it the\nSum of i, for x=1 to 2.5 (350,000e^-0.84x)\nNow this is not easily solvable, now I will assume that you have learned integrals, if not this is not going to make sense.  I do not remember the priciple name for what I am about to say, but we can take a finite sum and convert it to a definite integral, and the answer will be the same, so\ntake the definite integral from 1 to 2.5 of 350,000e^-0.84x dx\nThis is (350,000*e^-0.84x)/-0.84 from 1 to 2.5 (see source for how I solved this)\nSo following the rules for definite integrals we get,\np=(350,000*e^(-0.84*2.5)/-0.84)-(350,000*e^(-0.84*1)/-0.84)\n =((350,000*e^-2.1)/-0.84)-(350,000*e^-0.84)/-0.84)\n =(350,000*0.1225/-0.84)-(350,000*0.4317/-0.84)\n =(42859.7499/-0.84)-(151098.6832/-0.84)\n =-51,024-(-179,880)=230,904\nAssuming my math is right, which I now doubt, that's the answer I get, you might want to double check my integration using proper substitution, if you get the same answer turn it in as that
957350
x= 5/4 or x= -5/4 if\n\nx=(25^1/2 +x)/5,   5x=25^1/2+x,   4x=+ or - 5,   x=+ or - 5/4\n\nx= 6 1/4 or x= -6 1/4 if\n\nx=25^1/2 + x/5,  4/5x=+ or - 5,  x=+ or - 25/4