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Challenging Trig Expression Integration

Date: 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:


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


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 
Associated Topics:
College Calculus

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