What is $\int \frac{4\sin 8x}{\sqrt{1-\cos^4 4x}}dx$?

$\sin^{-1}(cos^24x)+C$ $-\sin^{-1}(cos^24x)+C$ $-\cos^{-1}(cos^22x)+C$ $-\sin^{-1}(cos 4x)+C$

$-\sin^{-1}(cos^24x)+C$

$I+\int\frac{sin8x}{\sqrt{1-cos^44x}}dx$ Let $\cos^2 4x=t$. Differentiating, $dt=(2cos4x).(-4sin4x) dx$ $dt=-4sin8xdx$.....(1) $I+\int-\frac{1}{4}.\frac{dt}{\sqrt{1-t^2}}$ $=-\frac{1}{4}.sin^{-1}(t)+C$ $=-\frac{1}{4}sin^{-1}(cos^24x)+C$