Question

Высшая математика, неопределенный интеграл

Answer 1

$\int\ (cos^4\frac{x}{2} sin^4\frac{x}{2})}dx=\int\ (\frac{1}{2}sinx)^4dx=\frac{1}{16}\int sin^4xdx=\frac{1}{16}\int (\frac{1-cos2x}{2})^2 dx=\\\\=\frac{1}{64}\int(1-cos2x)^2dx=\frac{1}{64}\int(1-2cos2x+cos^22x)dx=\frac{1}{64}x-\frac{1}{64}sin2x+\\\\+\frac{1}{64}\int \frac{1+cos4x}{2}dx= \frac{1}{64}x-\frac{1}{64}sin2x+\frac{1}{128}x+\frac{1}{128}\int cos4xdx=\\\\= \frac{3}{128}x-\frac{1}{64}sin2x+\frac{1}{512}sin4x+C$