solve $\int_0^1\cos x^{1/3} dx$ solve $$\int_0^1\cos x^{1/3} dx$$
My approach would be to use substitution and set $\cos x =t, dx = - \sin (t) dt$ which gives me $-\int_0^1t^{1/3} \sin( t) dt$ however then I'm stuck again...

For $n>-\dfrac12$ we have $\displaystyle\int_0^\frac\pi2\cos^{2n}x=\int_0^\frac\pi2\sin^{2n}x=\frac12\cdot B\Big(n+\tfrac12~;~\tfrac12\Big)$ . See Wallis' integrals for more details on their relationship to the Beta and $\Gamma$ functions. If however you are positive that the upper limit is indeed $1$ instead of $\dfrac\pi2$, then the result can only be expressed in terms of an incomplete beta function, $\displaystyle\int_0^a\cos^{2n}x=\frac12\bigg[B\Big(n+\tfrac12~;~\tfrac12\Big)-B\Big(\cos^2a~;~n+\tfrac12~;~\tfrac12\Big)\bigg]$. In this particular case, $n=\dfrac16$ and $a=1$.
