Question

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Answer 1

$\int sin^4x\, dx=\int (sin^2x)^2dx=\int \Big (\frac{1-cos2x}{2}\Big )^2dx=\\\\=\frac{1}{4}\cdot \int (1-2cos2x+cos^22x)dx=\frac{1}{4}\cdot \int (1-2cos2x+\frac{1+cos4x}{2})dx=\\\\=\frac{1}{4}\cdot \int (\frac{3}{2}-2cos2x+ \frac{1}{2}\cdot cos4x)dx=\\\\=\frac{1}{4}\cdot (\frac{3}{2}\cdot x-2\cdot \frac{1}{2}\cdot sin2x+\frac{1}{2}\cdot \frac{1}{4}\cdot sin4x)+C=\\\\=\frac{3}{8}\cdot x-\frac{1}{4}\cdot sin2x+\frac{1}{32}\cdot sin4x)+C$

$\int\limits^{\frac{\pi}{3}}_0 sin^4x\, dx=(\frac{3}{8}x-\frac{1}{4}\cdot sin2x+\frac{1}{32}\cdot sin4x )\Big |_0^{\frac{\pi}{3}}=\\\\= \frac{\pi}{8}-\frac{1}{4}\cdot sin\frac{2\pi}{3}+\frac{1}{32}\cdot sin\frac{4\pi }{3}=\frac{\pi}{8}-\frac{\sqrt3}{8}-\frac{\sqrt3}{64}=\frac{\pi }{8}- \frac{9\sqrt3}{64}$