Question

Помогите решить неопределенный интеграл
Int (1+sin^4 x) dx

Answer 1

$\int (1+sin^4x)dx=\int dx+\frac{1}{4}\int(1-cos2x)^2dx=\int dx+\\+\frac{1}{4}\int(1-2cos2x+cos^22x)dx=\int dx+\frac{1}{4}\int dx-\frac{1}{4}\int cos2xd(2x)+\\+\frac{1}{8}\int(1+cos4x)dx=\frac{11}{8}\int dx-\frac{1}{4}\int cos2xd(2x)+\frac{1}{32}\int cos4xd(4x)=\\=\frac{11x}{8}-\frac{sin2x}{4}+\frac{sin4x}{32}+C$