Question

IntegralSin4x× sin 8x dx

Answer 1

Recall that


`\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y`

This means


`\cos12x=\cos4x\cos8x-\sin4x\sin8x`

`\cos4x=\cos4x\cos8x+\sin4x\sin8x`

which then means


`\sin4x\sin8x=\frac{\cos4x-\cos12x}2`

So,


`\displaystyle\int\sin4x\sin8x\,\mathrm dx=\frac12\int(\cos4x-\cos12x)\,\mathrm dx`

`=\frac12\left(\frac14\sin4x-\frac1{12}\sin12x\right)+C`

`=\frac18\sin4x-\frac1{24}\sin12x+C`