Evaluate $\int_{-2}^2 (2x^2-4)^4dx$ Evaluate $\int_{-2}^2 (2x^2-4)^4dx$ So, I was trying to solve, I tried to expand this and got something, 347?. Is there like any other way to find the definite integral of this? Because after I expanded to raised to the $4th$ power it took a lot of my time. I was just wondering if there's like an easier way to do it? 
 A: Using symmetry:
$$\int_{-2}^2 (2x^2 - 4)^4\;\mathrm dx = 2 \int_0^2 (2x^2 - 4)^4 \;\mathrm dx = 32 \int_0^2 (x^2 - 2)^4 \;\mathrm dx$$
Then $(a+b)^4 = a^4 + 4 a^3b + 6a^2b^2 + 4ab^3 + b^4$ by the binomial thm.
$$(x^2-2)^4 = x^8 -8x^6 +24x^4 -32x^2+16$$
This results in
$$I = \frac{109568}{315}\approx 347.83$$
A: You could make it a little easier by noting that the integrand is even and factoring out a $2$, so
$$I=2\times2^4\int_0^2 (x^2-2)^4\,dx\ .$$
Apart from that, the quickest way to do the integral is to practise your algebra.  Hope that doesn't sound very unsympathetic, but really, if you expand by the binomial theorem you should be able to do most of this in your head.


*

*First term is $x^8$, integrated is $\frac19x^9$, substituting $2$ gives $\frac192^9$.

*Second term is $4(x^2)^3(-2)=-8x^6$, integrated is $-\frac172^3x^7$, substituting $2$ gives $-\frac172^{10}$.

*Three more terms, leave them to you.


So
$$I=2^5\Bigl(\frac192^9-\frac172^{10}+\cdots\Bigr)\ .$$
Looks like there will be lots of powers of $2$ to factor out.  The final calculation with the fractions will be a bit messy, but that's life.
