Evaluate the following integrals:
$\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} 4 x}} d x$
Let $t=\sin ^{2} 4 x$
$d t=2 \sin 4 x \cos 4 x \times 4 d x$
we know $\sin 2 x=2 \sin 2 x \cos 2 x$
therefore, $d t=4 \sin 8 x d x$
or, $\sin 8 x d x=d t / 4$
$\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} x}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{3^{2}+t^{2}}}$
Since we have, $\left.\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}\right.}+a^{2}\right)\right]+c$
$=\frac{1}{4} \int \frac{d t}{\sqrt{3^{2}+t^{2}}}=\frac{1}{4} \log \left[t+\sqrt{t^{2}+3^{2}}+c\right.$
$=\frac{1}{4} \log \left[\sin ^{2} 4 x+\sqrt{9+\sin ^{4} 4 x}+c\right.$
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