Integrate cos(x^5): Solve \int \cos x^5 dx

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SUMMARY

The integral \(\int \cos x^5 \, dx\) cannot be expressed in terms of elementary functions. Mathematica provides two forms for this integral: one involving the Exponential Integral function \(E(v,z)\) and another using the Gamma function \(\Gamma\). The first form is \(-\frac{1}{10} \left( E\left( \frac{4}{5}, -i x^5 \right) + E\left( \frac{4}{5}, i x^5 \right) \right)\), while the second form is \(-\frac{\left( x^{10} \right)^{\frac{4}{5}}}{10x^9}\left( \left(i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5}, -i x^5 \right) + \left(-i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5}, i x^5 \right) \right)\). The discussion highlights the convenience of using Taylor series for integration, suggesting a substitution of \(u = x^5\) for a more straightforward solution.

PREREQUISITES
  • Understanding of integral calculus and non-elementary integrals
  • Familiarity with the Exponential Integral function \(E(v,z)\)
  • Knowledge of the Gamma function \(\Gamma\)
  • Basic concepts of Taylor series and their applications in integration
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  • Study the properties and applications of the Exponential Integral function \(E(v,z)\)
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Artaxerxes
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Can you solve this \int \cos x^5 dx ?
 
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No, I can't
 
nope ...
 
It probably can't be done in terms of elementary functions. Mathematica has an answer of 2 different forms:

Take E(v,z) here to be this function http://functions.wolfram.com/GammaBetaErf/ExpIntegralE/ we have:\int \cos x^5 dx = - \frac{1}{10} \left( E\left( \frac{4}{5} , -i x^5 \right) + E\left( \frac{4}{5} , i x^5 \right) \right)

Or in terms of the gamma function http://functions.wolfram.com/GammaBetaErf/Gamma/ :

\int \cos x^5 dx = - \frac{\left( x^{10} \right)^{\frac{4}{5}}}{10x^9}\left( \left(i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , -i x^5 \right) + \left(-i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , i x^5 \right) \right)

I don't really understand why mathematica didn't simplify it further, so I've tried to keep it to what mathematica outputted.
 
Last edited:
Thank you!
 
thats why series solutions are so conviennt
 
The Taylor series for cos converges for all real x, so you should have just let x^5 = u in the taylor expansion of cos u, integrated term by term and you are left with an even nicer result than what mathematicia gives out.
 

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