Question

What is the antiderivative of `Cos(2x)Sin(x)dx`?

Answer 1

`intcos2xsinxdx=1/2int(2cos2xsinx)dx`
`=1/2int{sin(2x+x)-sin(2x-x)}dx`
`=1/2int(sin3x-sinx)dx`
`=1/2{-cos(3x)/3-(-cosx)}`
`=-1/6(cos3x-3cosx)+C.`

Answer 2

`int \ cos(2x)sinx \ dx = 1/2cosx-1/6cos3x+ C`

Explanation:

We use a little trick to express the integrand as the sum of multiple angles and then use the trig multiple angle formula:

`2 sin A cos B = sin (A +B) + sin (A -B)`

And we get;

`int \ cos(2x)sinx \ dx = 1/2 \ int \ 2sinxcos(2x) \ dx`
`" "= 1/2 \ int \ sin(x+2x) + sin(x-2x) \ dx`
`" "= 1/2 \ int \ sin(3x) + sin(-x) \ dx`
`" "= 1/2 \ int \ sin(3x) - sin(x) \ dx`

We can now easily integrate this:

`int \ cos(2x)sinx \ dx = 1/2 \ (-1/3cos3x+cosx) + C`
`" "= 1/2cosx-1/6cos3x+ C`