int sin theta sin2theta sin3thetasinθsin2θsin3θ ?

Evalute:
int sin theta sin2theta sin3thetasinθsin2θsin3θ

1 Answer
Jul 29, 2018

int(sinthetasin2thetasin3theta)d theta=cos(6x)/24-cos(4x)/16-cos(2x)/8+C(sinθsin2θsin3θ)dθ=cos(6x)24cos(4x)16cos(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)]sinαcosβ=12[sin(αβ)+sin(α+β)]
sinalphasinbeta=1/2[cos(alpha-beta)-cos(alpha+beta)]sinαsinβ=12[cos(αβ)cos(α+β)]
sin2alpha=2sinalphacosalphasin2α=2sinαcosα

So, let's use them. You can pick any pair. I picked sinthetasin2thetasinθsin2θ first:

int(sinthetasin2theta)sin3thetad theta(sinθsin2θ)sin3θdθ
=int(1/2)(costheta+cos3theta)(sin3theta)d theta=(12)(cosθ+cos3θ)(sin3θ)dθ
=(1/2)intsin3thetacostheta-sin3thetacos3thetad theta=(12)sin3θcosθsin3θcos3θdθ
=(1/2)int(1/2)(sin4theta+sin2theta)-(1/2)sin6thetad theta=(12)(12)(sin4θ+sin2θ)(12)sin6θdθ
=(1/4)int(sin4theta+sin2theta-sin6theta)d theta=(14)(sin4θ+sin2θsin6θ)dθ

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

(1/4)[-cos(4theta)/4-cos(2theta)/2+cos(6theta)/6]+C(14)[cos(4θ)4cos(2θ)2+cos(6θ)6]+C

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