How to integrate $\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx$ How do I integrate
\begin{equation}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx,
\end{equation}
which has arisen from a problem I'm working on? I've noticed I can do the following:
\begin{align}
\int_0^1\sqrt{-x^6+x^4-x^2+1}\:dx & =\int_0^1\sqrt{\left(-x^4\right)\left(x^2-1\right)-\left(x^2-1\right)}\:dx \\[3ex]
& = \int_0^1\sqrt{\left(-x^4-1\right)\left(x^2-1\right)}\:dx \\[3ex]
& = i\int_0^1 \sqrt{\left(x^4+1\right)\left(x+1\right)\left(x-1\right)}\:dx
\end{align}
But where do I go from here? Also, I'm a bit unsure about my last step above, i.e. not sure if it would be the right route to take.
Thanks in advance!
 A: $\int_0^1\sqrt{-x^6+x^4-x^2+1}~dx$
$=\int_0^1\sqrt{x^4(1-x^2)+1-x^2}~dx$
$=\int_0^1\sqrt{(1-x^2)(x^4+1)}~dx$
$=\int_0^\frac{\pi}{2}\sqrt{(1-\sin^2x)(\sin^4x+1)}~d(\sin x)$
$=\int_0^\frac{\pi}{2}(\cos^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}(1-\sin^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{4n}x}{4^n(n!)^2(1-2n)}dx-\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{4n+2}x}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(4n)!\pi}{2^{6n+1}(n!)^2(2n)!(1-2n)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(4n+2)!\pi}{2^{6n+3}(n!)^2(2n+1)!(1-4n^2)}$ (according to http://en.wikipedia.org/wiki/Wallis%27_integrals#Recurrence_relation.2C_evaluating_the_Wallis.27_integrals)
A: I think I can simplify it some, but I can't run it to ground.
Let $\theta = \cos^{-1} x$, so $x = \cos \theta$ and $dx = -\sin \theta \:d\theta$.  Also $1 - x^2 = 1 - \cos^2 \theta = \sin^2 \theta$.
$$\begin{align}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx & =
\int_{\frac{\pi}2}^0 -\sin \theta\sqrt{-\cos^6 \theta + \cos^4 \theta - \cos^2 \theta + 1} \:d\theta\\
& = \int_0^{\frac{\pi}2} \sin \theta\sqrt{(\cos^4 \theta + 1)(1 - \cos^2 \theta)} \:d\theta\\
& = \int_0^{\frac{\pi}2} \sin^2 \theta\sqrt{\cos^4 \theta + 1} \:d\theta\\
\end{align}$$
