Challenging Integral: Solving for P(cos(x)^3)/(x^6+1)

[joao_pimentel]

Hi guys

I'm wondering how shall I solve this primitive/integral

\int\frac{\cos^3 x}{x^6+1}dx

I really don't know how to start... which substitution shall I take or if I shall solve it by parts...

Any hints or tips will be very appreciated

Thanks in advance

 

[JJacquelin]

http://www.wolframalpha.com/input/?...ate+cos(x)^3/(x^6+1)*dx&x=4&fp=1&incTime=true
Are you sure that you really need an explicit expression of the integral ?

 

[joao_pimentel]

I don't know if I need an explicit expression, I just need to find how shall I start...

I've tried all trigonometric substitutions and nothing :(

I'm lost

 

[chiro]

Hey joao_pimentel.

Have you looked at different tables of integrals? Chances are you will find a table out there somewhere that gives a form that is exact or 'close' to the form you are working with.

One strategy I have for you is to reduce the trig terms to something that is at least linear. You might have to do a few integrations by parts by differentiating the trig terms and then using trig conversions to put them in the appropriate form.

The denominator of the term looks like a (1 + x^2) which is used in many tables, although if its in combination with a trig term then that could still make it complicated.

 

[chiro]

Also having looked at the solution, maybe another suggestion is to consider a transformation to an expontential form (in the form of using expressions of e^(whatever)) so for example cos(x) = 1/2(e^(ix) - e^(ix)).

 

[JJacquelin]

Split the integral in two, in writting cos(x)^3 in terms of cos(x) and cos(3x)
Split both integrals in six, in writting 1/((x^6)+1) in terms of a sum of six terms c/(x-r) where each r is on the form r=(-1)^(1/6) , i.e. the six complex roots of unit.
Then, the antiderivatives of each cos(x)/(x-r) and cos(3x)/(x-r) can be formally expressed in terms of the special functions Ci(z) and Si(z) where z is a complex linear function of x.

 

[D H]

Split both integrals in six, in writting 1/((x^6)+1) in terms of a sum of six terms c/(x-r) where each r is on the form r=(-1)^(1/6) , i.e. the six complex roots of unit.

That is the factorization of x⁶-1, not x⁶+1. The latter factorization is a bit uglier.

First factor x^6+1 as (x^2+1)(x^4-x^2+1). The latter term in turn has the factorization (x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1}), so

x^6+1 = (x^2+1)(x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})

I'll leave the last ugly bit as an exercise for the OP.

 

[joao_pimentel]

Thanks a lot guys... it was a big help...

I shall write cos(x)^3 in terms of a sum of cos(x) and cos(3x) as JJacquelin said and then fatorize x^6+1 as D H said... Then I shall use Ci(z) e Si(z) functions for each term..

I'll try it :)

Thanks a lot again

João

 

[joao_pimentel]

Guys

I know that cos^3(x)=\frac{1}{4}\left(cos(3x)+3cos(x)\right)

Then

\begin{aligned}<br /> \int &amp;\frac{\cos^3 x} {(x^2+1) (x^2-1+\sqrt[3]{-1}) (x^2-\sqrt[3]{-1})} dx \\<br /> &amp; =\frac{1}{4}<br /> \int \frac{\cos{3x}} {(x^2+1) (x^2-1+\sqrt[3]{-1}) (x^2-\sqrt[3]{-1})} dx +<br /> \frac{3}{4}<br /> \int \frac {\cos{x}} {(x^2+1)(x^2-1+\sqrt[3]{-1} )(x^2-\sqrt[3]{-1})} dx<br /> \end{aligned}<br />

Then I need to factorize \frac{1}{(x^2+1)(x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})} in partial fractions

<br /> \frac{1} {(x^2+1) (x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})} =<br /> \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2-1+\sqrt[3]{-1})}<br /> +\frac{Ex+F}{x^2-\sqrt[3]{-1})}<br />

Then I get each fraction \cos(a x)\frac{\alpha x + \beta}{x^2+\lambda}

How do I use such Ci(z) functions... Actually I confess I'm not aware of such Ci(z) and Si(z) functions... would it be Hyperbolic function?

Thank you a lot

 

[phyzguy]

Try this for an explanation of Ci(z) and Si(z):

http://en.wikipedia.org/wiki/Trigonometric_integrals

 

[JJacquelin]

I think that it not a good idea to process with partial fractons which denominators are second degree polynomials. This will lead to complicated integrals, not easy to expess in terms of Ci and Si functions.
I maintain what I already said in my previous post [ Y, 01:19 PM] :
1/((x^6)+1) is much easier to write as the sum of six terms on the form c/(x-r) where each c is a different complex number easy to find and where the six different r are the roots of (r^6)+1=0, i.e. six complex number easy to find.

 

[D H]

I maintain what I already said in my previous post [ Y, 01:19 PM]

Here's what you said yesterday:

Split both integrals in six, in writting 1/((x^6)+1) in terms of a sum of six terms c/(x-r) where each r is on the form r=(-1)^(1/6) , i.e. the six complex roots of unit.

That's incorrect. Using those (x-r) as the factors, where each r is one of the six sixth roots of unit, is the factorization of x^6-1=0, not x^6+1=0.

 

[JJacquelin]

Obviously :

the six different r are the roots of (r^6)+1=0, i.e. six complex number easy to find.

r = exp(i(2n+1)pi/6)
I think that it isn't the key point of the question.