Integral CalculusDate: 08/18/99 at 06:20:22 From: Kunal Kulkarni Subject: Integral calculus I don't know how to solve integral of sin^10x Date: 08/18/99 at 14:57:12 From: Doctor Rob Subject: Re: Integral calculus Thanks for writing to Ask Dr. Math. This can be done using integration by parts. INTEGRAL sin^n(x)*dx = INTEGRAL u*dv, = u*v - INTEGRAL v*du. Here you should pick u = sin^(n-1)(x) and v = -cos(x), so du = (n-1)*sin^(n-2)(x)*cos(x)*dx and dv = sin(x)*dx. Thus INTEGRAL sin^n(x)*dx = -sin^(n-1)(x)*cos(x) + (n-1)*INTEGRAL sin^(n-2)(x)*cos^2(x)*dx, = -sin^(n-1)(x)*cos(x) + (n-1)*INTEGRAL sin^(n-2)(x)*(1-sin^2(x))*dx, = -sin^(n-1)(x)*cos(x) + (n-1)*INTEGRAL sin^(n-2)(x)*dx - (n-1)*INTEGRAL sin^n(x)*dx. Now bring the least term on the right over to the left and divide by n: INTEGRAL sin^n(x)*dx = -(1/n)*sin^(n-1)(x)*cos(x) + ((n-1)/n)*INTEGRAL sin^(n-2)(x)*dx This formula lets you reduce the exponent from 10 to 8, then to 6, then to 4, then to 2, and finally to INTEGRAL dx, which is easy. Another approach is to use a trigonometric identity to express sin^10(x) as a sum of terms of the form a constant times sine or cosine of a multiple of x For example, sin^2(x) = (1/2)(1 - cos(2x)) = (1/2) - (1/2)cos(2x) sin^3(x) = (1/4)(3sin(x) - sin(3x)) = (3/4)sin(x) - (1/4)sin(3x) sin^4(x) = (1/8)(3 - 4cos(2x) + cos(4x)) = (3/8) - (1/2)cos(2x) + (1/8)cos(4x) and so on. These can be derived by using the identities sin(x)*sin(n*x) = (1/2)[cos([n-1]*x) - cos([n+1]*x)], sin(x)*cos(n*x) = (1/2)[sin([n-1]*x) + sin([n+1]*x)], repeatedly for different values of n. For example, to find sin^5(x), multiply the equation for sin^4(x) on both sides by sin(x), and use the second of these last two identities, twice: sin^5(x) = sin(x)*sin^4(x), = sin(x)*[(3/8) - (1/2)cos(2x) + (1/8)cos(4x)], = (3/8)sin(x) - (1/2)sin(x)cos(2x) + (1/8)sin(x)cos(4x) = (3/8)sin(x) - (1/2)(1/2)[sin([2-1]*x) + sin([2+1]*x)] + (1/8)(1/2)[sin([4-1]*x) + sin([4+1]*x)] = (3/8)sin(x) - (1/4)[sin(x) + sin([3x)] + (1/16)[sin(3x) + sin(5x)] = (3/8)sin(x) - (1/4)[sin(x) + sin(3x)] + (1/16)[sin(3x) + sin(5x)] = (3/8)sin(x) - (1/4)sin(x) - (1/4)sin(3x) + (1/16)sin(3x) + (1/16)sin(5x) = (1/8)sin(x) - (3/16)sin(3x) + (1/16)sin(5x) Once you have expressed sin^10(x) that way, then the integration is easy. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 09/23/1999 at 04:25:29 From: kunal kulkarni Subject: Integral calculus Sir, Can you also please tell me the integral of 1/x*(1+x^4)^3/4? Date: 09/24/1999 at 10:00:31 From: Doctor Rob Subject: Re: Integral calculus Assuming that the expression above is the same as (1/x)*([1+x^4]^3)/4, and you are integrating with respect to x, the answer may be found by expanding the cube using the Binomial Theorem, INTEGRAL (1/x)*(1+3*x^4+3*x^8+x^12)/4 dx, and expanding the integrand into powers of x, INTEGRAL (1/4)*x^(-1) + (3/4)*x^3 + (3/4)*x^7 + (1/4)*x^11 dx. Now the integration is pretty easy, and you can finish. Another approach would be to let u = 1 + x^4, and so du = 4*x^3 dx, dx/x = du/(4*[u-1]), and then you want to find INTEGRAL u^3/(4*[u-1]) du. This you can do by dividing u - 1 into u^3 by long division, and finding the quotient and remainder, and then integrating the resulting expression. This is also pretty easy, and you can do that one yourself, too. If instead you meant that the expression to be integrated to be (1/x)*(1+x^4)^(3/4), then this is one of those functions that cannot be integrated in closed form in terms of the familiar functions of calculus. My version of Mathematica(TM) gives the result: (1+x^4)^(3/4)/3 - Hypergeometric2F1[1/4,1/4,5/4,-x^(-4)]/Abs[x]. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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