#int sin theta sin2theta sin3theta# ?

Evalute:
#int sin theta sin2theta sin3theta#

1 Answer
Jul 29, 2018

#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#