Question

`int sin theta sin2theta sin3theta` ?

Evalute:
`int sin theta sin2theta sin3theta`

Answer 1

`int(sinthetasin2thetasin3theta)d theta=cos(6x)/24-cos(4x)/16-cos(2x)/8+C`

Explanation:

There's a few ways to tackle this problem, but personally I would expand the integrand by using various trigonometric identities, including the product-to-sum identities and double-angle formulas.

Here are the ones I used specifically for this problem:
`sinalphacosbeta=1/2[sin(alpha-beta)+sin(alpha+beta)]`
`sinalphasinbeta=1/2[cos(alpha-beta)-cos(alpha+beta)]`
`sin2alpha=2sinalphacosalpha`

So, let's use them. You can pick any pair. I picked `sinthetasin2theta` first:

`int(sinthetasin2theta)sin3thetad theta`
`=int(1/2)(costheta+cos3theta)(sin3theta)d theta`
`=(1/2)intsin3thetacostheta-sin3thetacos3thetad theta`
`=(1/2)int(1/2)(sin4theta+sin2theta)-(1/2)sin6thetad theta`
`=(1/4)int(sin4theta+sin2theta-sin6theta)d theta`

At this point, it's a simple indefinite integral:

`(1/4)[-cos(4theta)/4-cos(2theta)/2+cos(6theta)/6]+C`

`=cos(6x)/24-cos(4x)/16-cos(2x)/8+C`