Challenging Trig Expression IntegrationDate: 04/12/2008 at 11:07:36 From: Ryan Subject: How to integrate [(sinx)^3] / [(sinx)^3)+(cosx)^3)] For the life of me, I can't figure out how to integrate this function: (sinx)^3 --------------------- [(sinx)^3]+[(cosx)^3] There is no convenient form that this can take so that substitution, integration by parts, or both can be performed. I've looked in several textbooks and online sites and haven't seen anything remotely close to this, but the answer is exact and symbolic (no Taylor series needed). answer:(x/2) - (1/6)ln(cosx + sinx) + (1/3)ln(sin(2x)-2) I've tried a huge number of different trig identities to force this into a form which is integratable by substitution or by parts (probably need both). I also completed the square on several of them to make two pieces that are perhaps then able to be converted to integratable forms. Nothing works. I got the answer from Wolfram Integrator. I've commonly found that it can be misleading because a more human approach will produce an equivalent form that isn't necessarily obvious that it matches Wolfram's answer. The first part, (x/2), bothers me because every integration I've performed with trig functions produces a trig function or an exponential. So i used that as my starting point to try and figure out how that can even come about from a trig function, but so far I've gotten nowhere with that. Maybe this needs to be performed in complex or hypergeometric? Date: 04/13/2008 at 00:14:04 From: Doctor Ali Subject: Re: How to integrate [(sinx)^3] / [(sinx)^3)+(cosx)^3)] Hi Ryan! Thanks for writing to Dr. Math. It sounds like you've done a lot of good thinking so far. I think the trick to this is doing some algebraic manipulation. I'd start it this way: - | SIN(x)^3 | --------------------- dx | SIN(x)^3 + COS(x)^3 - Divide the top and bottom by COS(x)^3: - | TAN(x)^3 | -------------- dx | TAN(x)^3 + 1 - Assume TAN(x) = u, so (1 + TAN(x)^2) dx = du - | u^3 | ------------------- du | (u^3 + 1)(u^2 + 1) - Does that make sense? Let's try to break that fraction into an equivalent sum of fractions. This is the key algebraic step I referred to above: - | / 2u - 1 u - 1 1 \ | |--------------- - ---------- - -------- | du | \ 3(u^2 - u + 1) 2(u^2 + 1) 6(u + 1) / - Can you do this yourself? This way, we change the difficult integral to three not so difficult integrals. Hints: 1) The first integral can be solved using LN, because the top is a multiple of the bottom's derivative. 2) The second integral splits to give one LN and one ARCTAN. 3) The third integral is a simple LN. Can you continue the solution yourself? By the way, don't forget to substitute TAN(x) back for u at the end. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/