How to integrate $\int_{-3}^3 (x^2-3)^{3} \,dx$ without expanding the polynomial? 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.
 A: Perhaps more than a bit of overkill, but you could get a generating function for all the integrals $J_n = \int_{-3}^3 (x^2-3)^n \; dx$:
$$\eqalign{g(t) &= \sum_{n=0}^\infty t^n J_n\cr & = \int_{-3}^3 \sum_{n=0}^\infty t^n (x^2-3)^n\; dx \cr &= \int_{-3}^3 \dfrac{dx}{1 + 3 t - t x^2}\cr
&= \dfrac{2}{\sqrt{3t^2+t}} \text{arctanh}\left( \dfrac{3\sqrt{t}}{\sqrt{3t+1}}\right)\cr}$$
And then taking the first terms of the Taylor series:
$$g(t) = 6 + \dfrac{216}{5} t^2 + \dfrac{2592}{35} t^3 + \ldots $$
so that $J_3 = 2592/35$.
If that last is too much like an "expansion": $g(t)$ satisfies the differential equation
$$ 72 t g(t) + (144 t^2 + 12 t - 3) g'(t) + (36 t^3 + 6 t^2 - 2 t) g''(t) = 0$$
from which we can obtain a recursion for the coefficients $J_n$:
$$ \left( 36\,n+36 \right) J_n + \left( 6\,n+6 \right) J_{n+1} + \left( -2\,n-5 \right) J_{n+2} = 0$$ 
Given that $J_0 = 6$ and $J_1 = 0$ (which are easy to calculate), 
the case $n=0$ gives us $216 - 5 J_2 = 0$, so $J_2 = 216/5$, and then 
$n=1$ gives us $2592/5 - 7 J_3 = 0$, or $J_3 = 2592/35$.
A: $$\int_{-3}^{3}(x^2-3)^3\,dx  = 81 \int_{-1}^{1}(3u^2-1)^3\,du \stackrel{IBP}{=}81\left(16-18\int_{-1}^{1}x^2(3x^2-1)^2\,dx\right)\tag{1}$$
Now we may exploit the fact that the last integral is a $L^2$ norm. Since, in terms of Legendre polynomials:
$$ x(3x^2-1) = \frac{4}{5}\,P_1(x) + \frac{6}{5}\,P_3(x) \tag{2}$$
we have:
$$ \int_{-1}^{1} x^2(3x^2-1)^2\,dx = \frac{16}{25}\cdot\frac{2}{3}+\frac{36}{25}\cdot\frac{2}{7} = \frac{88}{105}\tag{3}$$
and the original integral equals $\color{red}{\large\frac{2592}{35}}$. Probably $(2)$ counts as an expansion, but it is a rather easy one.
A: 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 $I(1) = \frac{2592}{35}$.
