Question

Помогите решить интеграл $\int\limits {sin^6x*cos^4x} \, dx$, ответ должен быть: $\frac{1}{128}(\frac{3x}{2} - \frac{sin4x}{2} + \frac{sin8x}{16} ) - \frac{sin^52x}{320} + C$ (увеличила полученные баллы за ответ)

Answer 1

Ответ:

$\displaystyle \int sin^6x\cdot cos^4x\, dx=\int \Big((sinx\cdot cosx)^2\Big)^2\cdot sin^2x\, dx=\int \Big(\frac{sin^22x}{4}\Big)^2\cdot \frac{1-cos2x}{2}\, dx=\\\\\\=\frac{1}{32}\int \frac{(1-cos4x)^2}{2^2}\cdot (1-cos2x)\, dx=\\\\\\=\frac{1}{128}\int \Big(1-2cos4x+cos^24x\Big)\Big(1-cos2x\Big)\, dx=\\\\\\=\frac{1}{128}\int \Big(1-cos2x-2cos4x+2\, cos4x\cdot cos2x+\frac{1+cos8x}{2}-cos^24x\cdot cos2x\Big)\, dx=\\\\\\=\frac{1}{128}\int \Big(\frac{3}{2}-cos2x-2cos4x+(cos6x+cos2x)+\frac{1}{2}cos8x\Big)-$

$\displaystyle -\frac{1}{128}\int \frac{1+cos8x}{2}\cdot cos2x\, dx=\frac{1}{128}\cdot \Big(\frac{3}{2}\, x-\frac{1}{2}sin4x+\frac{1}{6}\, sin6x+\frac{1}{16}\, sin8x\Big)-\\\\\\-\frac{1}{128\cdot 2}\int \ig(cos2x+cos8x\cdot cos2x)\, dx=\\\\\\=\frac{1}{128}\cdot \Big(\frac{3}{2}\, x-\frac{1}{2}sin4x+\frac{1}{6}\, sin6x+\frac{1}{16}\, sin8x\Big)-\\\\\\-\frac{1}{256}\int \Big(cos2x+\frac{1}{2}\, (cos10x+cos6x)\Big)\, dx=$

$\displaystyle =\frac{1}{128}\cdot \Big(\frac{3}{2}\, x-\frac{1}{2}sin4x+\frac{1}{6}\, sin6x+\frac{1}{16}\, sin8x\Big)-\\\\\\-\frac{1}{512}\, sin2x-\frac{1}{5120}\, sin10x-\frac{1}{3072}\, sin6x+C$