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!


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}


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