i need an answer to a maths integration question i will use (int) for the integration sign (^) for the power?
969285
It's ambiguous because you used no spaces or brackets.\nWas that "1/2 X+5" or "1/(2X+5)"?\n\nSince (int) (1/2 X+5)^6 dx is pretty easy as a Binomial Theorem expansion or substitution (Don't forget the constant of integration as person below did!), presume you meant the other possibility:\n\n(int) (1/(2X+5))^6 dx = (int) (2X+5)^-6 dx\nIntegrate by substituting (and possibly by parts): \nw = 2x+5, dw = 2dx\n\n(int) (2X+5)^-6 dx = 1/2 (int) w^-6 dw\n= (-1/2*7) w^-7 + C\n= (-1/14) w^-7 + C\n\n(Remember the constant of integration.)
133829
8x^7\n\nThe first reference refers to this as the "Power Rule".\n\n  x^n'=nx^{n-1}
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
981411
We use the symbol above the number 6 on the keyboard to show exponents.  For example, 2 squared is typed 2^2 = 4.\n\nWhy don't you use the symbol ^ and retype your question correctly?\n\nGuido
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
408042
Huh? To 3sqrx2 1? You'll need to give me more details, your question doesn't make sense otherwise.\n\nAnd use ^ as an exponent. I'm not sure if it's (8x^2 + 2x - 2) or (8x^2 + 2x^-2).\n\nAnyway, the indefinite integral of 8x^2 + 2x - 2 is (8x^3)/3 + x^2 - 2x. The indefinite integral of 8x^2 + 2x^-2 is (8x^3)/3 - 2/x. I can't get you any farther than that, because the rest of your question is incomprehensible.
1148067
You elevate the number to the power 2 like this:\n\n                                    x^2\n\nYou should still do the normal product but it is easier to thing you et the square of a certain number and most calculators already have the elevate to the second power function integrated\n\nCynthia
330547
#include<math.h>\nthen, pow(10,i)\nwill give the antilog of i.
200165
I will assume cos(x)^3 means (cos x)^3 and does NOT mean  cos(x^3)\nFirst start by separating the integral into two pieces.\nI will use "int" to stand for the integral sign.\nint{(cosx)^3/(cosx)^2}dx - int{(sinx)^2/(cosx)^2}dx\n\nNow (cosx)^3/(cosx)^2 = cosx\nand (sinx)^2/(cosx)^2 = (tanx)^2 = (secx)^2 - 1\nThus, the integral is now equal to \n\nint{cosx}dx - int{(secx)^2 - 1}dx\n\nThe antiderivative of cosx is sinx.\nThe antiderivative of (secx)^2 is tanx.\nThe antiderivative of -1 is -x.\n\nHence, the solution is \nsinx - tanx + x + C\n\n(The signs changed for tanx and -x because there was a negative sign in front of their integral, and the negative sign has to be distributed).
427075
If you took the 3 from the denominator, you'd have\n\n1/3 int{(x^2 + 2x - 2) / (sqrt(x)}\n\nsince you can move constants in and out of an integral with no problems.  If you got rid of the radical, you'd then have\n\n=1/3 int{(x^2 + 2x - 2) / x^(1/2)}\n=1/3 int{x^2/x^(1/2) + 2x/x^(1/2) - 2/ x^(1/2)}\n=1/3 int{x^(3/2) + 2x^(1/2) - 2x^(-1/2)}\n=1/3[(2/5)x^(5/2) + 2(2/3)x^(3/2) - 2(2)x^(1/2)]\n=(2/15)x^(5/2) + (4/9)x^(3/2) - (4/3)x^(1/2)\n\nEvaluating from 1 to 8 gives you\n\n=[(2/15) x (8)^(5/2) + (4/9) x (8)^(3/2) - (4/3) x (8)^(1/2)]\n-[(2/15) x (1)^(5/2) + (4/9) x (1)^(3/2) - (4/3) x (1)^(1/2)]\n\n= [256sqrt(2)/15 + 64sqrt(2)/9 - 8sqrt(2)/3] - [(2/15) + (4/9) - (4/3)]\n\n= 968sqrt(2)/45 - 34/45\n\n= (968sqrt(2)-34) / 45
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)]
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
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
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
721751
let theta be represented by 't' and integral sign be replaced by I. Then\n\nI[1 - cos(2t)dt\n= I[1*dt - cos(2t)*dt]\n= I(dt) - I[cos(2t)]dt\n= t - [sin(2t)]/2 + C ___________(1)\nwhere C = arbitrary constant of integration\n\nFrom (1)\nWhen t = 0, I(0) = 0 - 0 + C = C \nWhen t = T, I(T) = T - [sin(2T)]/2 + C\n\nFrom the Fundamental Theorem of Integral Calculus we can write that \nthe reqd. integral = I(T) - I(0)\n= T - [sin(2T)]/2 + C - C\n= T - [sin(2T)]/2\n= T - {sin(T)}*{cos(T)} [since sin(2T) = 2*{sin(T)}*{cos(T)}]
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
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