3

My integral is $\int \cos(x^{\frac{3}{4}})dx$. I did the substitution $u=x^{\frac{3}{4}}$ then $\frac{4}{3}u^{\frac{1}{3}}du=dx$ and my new integral is $$ \frac{4}{3}\int \cos(u)u^{\frac{1}{3}}du$$.

Now my intuition is to try with integration by parts, $a=u^{\frac{1}{3}}$ and $db= \cos(u) du$ then $da= \frac{1}{3}u^{-\frac{2}{3}}du$ and $b= \sin u$. Therefore,

$$\frac{4}{3}\int \cos(u)u^{\frac{1}{3}}du= \frac{4}{3}u^{\frac{1}{3}}\sin u -\frac{4}{9}\int u^{-\frac{2}{3}}\sin u du$$ but if do again by parts choosing all possibilities i get $0=0$ or i will never stop with integrals of type $\int \sin u u^{q}$ or $\int \cos u u^{q}$.

Does not the integral have primitive.? The only way is with complex numbers.? Or maybe i miss something? Thank you so much for hints or the solution!

Best

Anne Bauval
  • 34,650
weymar andres
  • 1,253
  • 9
  • 15

1 Answers1

2

Maple expresses it using Lommel functions:

$$ x \cos \! \left(x^{\frac{3}{4}}\right)+\frac{x^{\frac{3}{8}} \sin \! \left(x^{\frac{3}{4}}\right) s_{\frac{5}{6},\frac{3}{2}}^{\left(+\right)}\! \left(x^{\frac{3}{4}}\right)}{3}-x^{\frac{3}{8}} s_{\frac{11}{6},\frac{1}{2}}^{\left(+\right)}\! \left(x^{\frac{3}{4}}\right) \cos \! \left(x^{\frac{3}{4}}\right)+\frac{s_{\frac{11}{6},\frac{1}{2}}^{\left(+\right)}\! \left(x^{\frac{3}{4}}\right) \sin \! \left(x^{\frac{3}{4}}\right)}{x^{\frac{3}{8}}} $$ These can also be expressed using hypergeometric functions: $$ \frac{3 \left(-3 x^{\frac{5}{2}} \cos \! \left(x^{\frac{3}{4}}\right)+10 x^{\frac{7}{4}} \sin \! \left(x^{\frac{3}{4}}\right)\right) {}_{1}^{}{{{F_{2}^{}}}}\! \left(1;\frac{13}{6},\frac{8}{3};-\frac{x^{\frac{3}{2}}}{4}\right)}{70}-\frac{81 x^{\frac{13}{4}} \sin \! \left(x^{\frac{3}{4}}\right) {}_{1}^{}{{{F_{2}^{}}}}\! \left(2;\frac{19}{6},\frac{11}{3};-\frac{x^{\frac{3}{2}}}{4}\right)}{7280}+x \cos \! \left(x^{\frac{3}{4}}\right) $$

Robert Israel
  • 448,999