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I want to compute the integral $\int \sqrt[3]{1+\sin x}\ dx$ via Taylor series. My idea is : find Taylor expansion around zero of the function $f(x)= \sqrt[3]{1+\sin x}=\displaystyle\sum_{n=0}^{\infty} c_nx^n$, and after to integrate the Taylor expansion. Then $\int \sqrt[3]{1+\sin x}\ dx=\displaystyle\sum_{n=0}^{\infty}\frac{c_n}{n+1} x^{n+1}.$

Firt question: Am I right?

Second question: Is it difficult to find the Taylor expansion? I believe the way to find $f^{(n)}(0)$ for all $n$, is difficult in a sense of computing the derivatives.

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  • $\begingroup$ $\int\sqrt[3]{1+\sin x}\,dx$ is not a function but a class of functions. For the sake of accuracy, you should replace it with $\int_{0}^{x}\sqrt[3]{1+\sin t}\,dt$. $\endgroup$ Jul 12 '20 at 12:44
  • $\begingroup$ Yes I know that, that is I mean $\endgroup$ Jul 12 '20 at 13:06
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It's not difficult $-$ you don't need to compute the derivatives $-$ but it does get messy. Suppose we want the Taylor expansion up to $x^4$. First, the Taylor expansion of $\sqrt[3]{1+u}$ is

$$\sqrt[3]{1+u}=1+\frac13u-\frac19u^2+\frac{5}{81}u^3-\frac{10}{243}u^4+O(u^5)$$

Second, the Taylor expansion of $\sin x$ is

$$\sin x=x-\frac16x^3+O(x^5)$$

From this you get the powers of $\sin x$:

$$\sin^2 x=x^2-\frac13x^4+O(x^5)$$

$$\sin^3 x=x^3+O(x^5)$$ $$\sin^4 x=x^4+O(x^5)$$ And now you can substitute these expansions for $u, u^2,u^3,$ and $u^4$ to get your answer.

As you will appreciate, this becomes more and more complicated as you need more and more terms. But for small $x$, it will converge rapidly.

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