Answer: \(\sin^6\theta=\frac{5}{16}-\frac{15}{32}\cos2\theta+\frac{3}{16}\cos4\theta-\frac{1}{32}\cos6\theta.\)

\(\displaystyle \int_0^{\pi/4}\sin^6(2x)\,\mathrm dx=\frac{5\pi}{64}.\)

Using de Moivre's theorem or standard multiple-angle formulae,

\(\sin^6\theta=\left(\frac{1-\cos2\theta}{2}\right)^3.\)

Expanding and reducing powers of cosine gives

\(\sin^6\theta=\frac{5}{16}-\frac{15}{32}\cos2\theta+\frac{3}{16}\cos4\theta-\frac{1}{32}\cos6\theta.\)

Now put \(\theta=2x\):

\(\sin^6(2x)=\frac{5}{16}-\frac{15}{32}\cos4x+\frac{3}{16}\cos8x-\frac{1}{32}\cos12x.\)

Therefore

\(\int_0^{\pi/4}\sin^6(2x)\,\mathrm dx=\left[\frac{5x}{16}-\frac{15}{128}\sin4x+\frac{3}{128}\sin8x-\frac{1}{384}\sin12x\right]_0^{\pi/4}.\)

At \(x=\frac{\pi}{4}\), all sine terms are zero: \(\sin\pi=\sin2\pi=\sin3\pi=0\). The sine terms are also zero at \(x=0\). Hence

\(\int_0^{\pi/4}\sin^6(2x)\,\mathrm dx=\frac{5}{16}\cdot\frac{\pi}{4}=\frac{5\pi}{64}.\)