# Compute the $\int \sqrt[3]{1+\sin x}\ dx$ by help of Taylor series.

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.

• $\int\sqrt[3]{1+\sin x}\,dx$ is not a function but a class of functions. For the sake of accuracy, you should replace it with $\int_{0}^{x}\sqrt[3]{1+\sin t}\,dt$. Jul 12 '20 at 12:44
• Yes I know that, that is I mean Jul 12 '20 at 13:06

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.