Question

Решите задачу №9 пожалуйста

Answer 1

$\sin6x\sin4x=\frac12\cdot\left(\cos(6x-4x)-\cos(6x+4x)\right)=\frac12\cdot\left(\cos2x-\cos10x\right)\\\\\int\sin6x\sin4xdx=\frac12\int(\cos2x-\cos10x)dx=\frac12\int\cos2xdx-\frac12\int\cos10xdx=\\\\=\frac12\cdot\frac12\sin2x-\frac12\cdot\frac1{10}\sin10x+C=\frac14\sin2x-\frac1{20}\sin10x+C$