Решить интеграл sin^2x*cos^4x*dx

[latex]\int sin^2x\cdot cos^4x\, dx=\int \frac{1-cos2x}{2}\cdot (\frac{1+cos2x}{2})^2dx=\\\\=\frac{1}{8}\int (1-cos^22x)(1+cos2x)dx=\\\\=\frac{1}{8}\int (1-cos^22x+cos2x-cos^32x)dx=\\\\=\frac{1}{8}\int (1-\frac{1+cos4x}{2}+cos2x)dx-\frac{1}{8}\int cos^22x\cdot cos2xdx=\\\\=\frac{1}{8}\int (\frac{1}{2}-\frac{1}{2}cos4x+cos2x)dx-\frac{1}{8}\int (1-sin^22x)\cdot \frac{1}{2}d(sin2x)=\\\\=\frac{1}{8}(\frac{1}{2}x-\frac{1}{8}sin4x+\frac{1}{2}sin2x)-\frac{1}{16} (sin2x-\frac{sin^32x}{3})+C[/latex]