TITLE: indefinite integral $\int\sin\sqrt[3]{x}~dx$
QUESTION [6 upvotes]: I want to determinate the integral  $\int\sin\sqrt[3]{x}~dx$ .
I tried to use integration by partitions and integration by substitution but I came to no result. 
I know the result which is shown here http://www.wolframalpha.com/input/?i=sin%28root%28x%2C3%29%29+dx
but I want to know the steps and how to come there!

REPLY [12 votes]: Let $x = u^3$, and then integrate by parts to get rid of the polynomial.
Details:
\begin{align}
  \int \sin(x^{1/3}) dx &=\int \sin u (3u^2 du) \\
&=-3u^2\cos u  + \int 6u\cos u du\\
&= -3u^2\cos u  + 6u\sin u -\int 6\sin u du\\
&=-3u^2\cos u  + 6u\sin u + 6\cos u\\
\int \sin(x^{1/3}) dx &=
-3x^{2/3}\cos x^{1/3}  + 6x^{1/3}\sin x^{1/3} + 6\cos x^{1/3}
 \end{align}