Question

Найти определенный интеграл
Интеграл 2pi cos(x/4)^8 dx
0

Answer 1

$\int cos^8\frac{x}{4}\, dx=\Big[\; cos^2\alpha =\frac{1+cos2\alpha }{2}\; \Big]=\int \Big(\frac{1+cos\frac{x}{2}}{2}\Big)^4\, dx=\\\\=\frac{1}{16}\int \Big(1+4cos^3\frac{x}{2}+6cos^2\frac{x}{2}+4cos\frac{x}{2}+(cos^2\frac{x}{2})^2\Big)dx=\\\\=\frac{1}{16}\cdot \Big(x+4\int (1-sin^2\frac{x}{2})\cdot cos\frac{x}{2}\, dx+4\int cos\frac{x}{2}\, dx+\int \Big(\frac{1+cosx}{2}\Big)^2\Big)=$

$=\frac{x}{16}+\frac{1}{4}\cdot \Big (\int cos\frac{x}{2}\, dx-2\int sin^2\frac{x}{2}\cdot d(sin\frac{x}{2})\Big)+\frac{1}{2}\cdot sin\frac{x}{2}+\\\\+\frac{1}{16\cdot 4}\int (1+2cosx+cos^2x)\, dx=\frac{x}{16}+\frac{1}{2}\cdot sin\frac{x}{2}-\frac{1}{2}\cdot \frac{sin^3\frac{x}{2}}{3}+\frac{1}{2}\cdot sin\frac{x}{2}+\\\\+\frac{1}{64}\cdot (x+2sinx+\int \frac{1+cos2x}{2}\, dx)=$

$=\frac{x}{16}+sin\frac{x}{2}-\frac{1}{6}\cdot sin^3\frac{x}{2}+\frac{x}{64}+\frac{1}{32}\cdot sinx+\frac{1}{128}\, x+\frac{1}{256}\, sin2x+C\; ;\\\\\\\int \limits _0^{2\pi }cos^8\frac{x}{4}\, dx=(\frac{x}{16}+sin\frac{x}{2}-\frac{1}{6}\cdot sin^3\frac{x}{2}+\frac{x}{64}+\frac{1}{32}\cdot sinx+\frac{1}{128}\, x+\frac{1}{256}\, sin2x)\Big|_0^{2\pi }=\\\\=\frac{\pi}{8}+\frac{\pi }{32}+\frac{\pi}{64}=\frac{11\pi}{64}$