TITLE: How to integrate $\int_{-3}^3 (x^2-3)^{3} \,dx$ without expanding the polynomial?
QUESTION [5 upvotes]: How can I integrate:
$$\int_{-3}^3 (x^2-3)^{3} \,dx,$$
neither expanding the polynomial nor using the relationship between integral and derivatives? I mean, there is a way to compute this integral using for instance properties of integral? like the invariance by translation?
Thank you.

REPLY [5 votes]: For $\theta$ in $\mathbb{R}$ define $I(\theta) = \int_{-3}^{3}{({x^2-3\theta})^3 dx}$. Note that the integrand is a polynomial in $\theta$ of degree $3$ and because you are integrating in $x$, then $I(\theta)$ is a polynomial in $\theta$ of degree less than or equal to $3$. Because of this, $I(\theta)$ agrees with its Taylor polynomial of order $3$ centered at $0$. Differentiating under the integral sign, you can calculate
$$ I^{(n)} (0) = \frac{\partial ^ nI}{\partial \theta^n} \bigg|_{\theta = 0} = \int_{-3}^{3}{ \frac{\partial^n}{\partial \theta^n}   ({x^2-3\theta})^3  \bigg|_{\theta = 0}  dx}$$
Note that because you are evaluating at $\theta=0$ you will not have to expand powers. You get $$I(\theta) = \frac{2}{7}\cdot3^7 - \frac{2}{5} \cdot 3^7 \cdot\theta + 2 \cdot 3^5 \cdot\theta^2 - 2 \cdot 3^4 \cdot\theta^3$$
Now $I(1) = \frac{2592}{35}$.