Choosing Factors When Integrating by Parts
Date: 10/08/2003 at 21:23:52 From: Consuelo Subject: Integration by parts I use integration by parts to find 1 / | (r^3)/[(4+r^2)^(1/2)] dr / 0 What I find most difficult is finding u, v, du, and dv. I tried separating the integral by making one simple integral to take the anti-derivative of and then using the difficult integral to do integration by parts. I just don't know how to start it.
Date: 10/11/2003 at 13:53:48 From: Doctor Fenton Subject: Re: Integration by parts Hi Consuelo, Thanks for writing to Dr. Math. I like to write the integration by parts formula in the form / / | u v' dx = uv - | u' v dx , / / because it arises from the Product Rule for derivatives: [u v]' = u' v + u v' . The idea is to look at the integrand as a product of two functions, and one of them must be an exact derivative (or maybe a constant times an exact derivative). The trick is to choose the factors correctly. Formally, it looks as if you are "moving" the derivative from one factor to the other. This will help only if the new integrand is simpler in some sense than the original. In your example, the integrand is (r^3)/[(4+r^2)^(1/2)] The square root is a unit which you can't break up, so the only question is how to distribute the powers of r. Essentially, the only choices are to take the factors to be one of the following: r^3 and (4+r^2)^(-1/2) r^2 and r*(4+r^2)^(-1/2) r and r^2*(4+r^2)^(-1/2) Of course, we also have to choose one to be u (the factor to be differentiated) and one to be v' (the factor to be integrated). You can differentiate (4+r^2)^(-1/2) but if you do, you will obtain a factor of (4+r^2)^(-3/2) plus an additional power of r from applying the Chain Rule. You would also have to integrate r^3, which would produce r^4, so the integrand in the new integral after using this approach would be a constant times r^5*(4+r^2)^(-3/2) That looks worse than the original integral: it has more powers of r, and a larger power of (4+r^2) in the denominator. This observation strongly suggests that you want to integrate (4+r^2)^(-1/2). However, if you try to use the substitution u = 4+r^2 then du = 2rdr, so you need a power of r to integrate this term. That means that you should write the original integrand as the product of r^2 and r*(4+r^2)^(-1/2) taking u = r^2 and v' = r*(4+r^2)^(-1/2) . After integrating, you will have an integrand which is a constant times r*(4+r^2)^(1/2) and that can be evaluated with another integration by substitution. If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum