$\int sin θ sin 2 θ sin 3 θ$ $int sin theta sin2theta sin3theta$ ?

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$\int sin θ sin 2 θ sin 3 θ$ $int sin theta sin2theta sin3theta$ ?

Evalute:
$\int sin θ sin 2 θ sin 3 θ$ $int sin theta sin2theta sin3theta$

1 Answer

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Answer 1 · 2018-07-29T00:50:23

$\int (sin θ sin 2 θ sin 3 θ) d θ = \frac{cos (6 x)}{24} - \frac{cos (4 x)}{16} - \frac{cos (2 x)}{8} + C$ $int(sinthetasin2thetasin3theta)d theta=cos(6x)/24-cos(4x)/16-cos(2x)/8+C$

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:
$sin α cos β = \frac{1}{2} [sin (α - β) + sin (α + β)]$ $sinalphacosbeta=1/2[sin(alpha-beta)+sin(alpha+beta)]$
$sin α sin β = \frac{1}{2} [cos (α - β) - cos (α + β)]$ $sinalphasinbeta=1/2[cos(alpha-beta)-cos(alpha+beta)]$
$sin 2 α = 2 sin α cos α$ $sin2alpha=2sinalphacosalpha$

So, let's use them. You can pick any pair. I picked $sin θ sin 2 θ$ $sinthetasin2theta$ first:

$\int (sin θ sin 2 θ) sin 3 θ d θ$ $int(sinthetasin2theta)sin3thetad theta$
$= \int (\frac{1}{2}) (cos θ + cos 3 θ) (sin 3 θ) d θ$ $=int(1/2)(costheta+cos3theta)(sin3theta)d theta$
$= (\frac{1}{2}) \int sin 3 θ cos θ - sin 3 θ cos 3 θ d θ$ $=(1/2)intsin3thetacostheta-sin3thetacos3thetad theta$
$= (\frac{1}{2}) \int (\frac{1}{2}) (sin 4 θ + sin 2 θ) - (\frac{1}{2}) sin 6 θ d θ$ $=(1/2)int(1/2)(sin4theta+sin2theta)-(1/2)sin6thetad theta$
$= (\frac{1}{4}) \int (sin 4 θ + sin 2 θ - sin 6 θ) d θ$ $=(1/4)int(sin4theta+sin2theta-sin6theta)d theta$

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

$(\frac{1}{4}) [- \frac{cos (4 θ)}{4} - \frac{cos (2 θ)}{2} + \frac{cos (6 θ)}{6}] + C$ $(1/4)[-cos(4theta)/4-cos(2theta)/2+cos(6theta)/6]+C$

$= \frac{cos (6 x)}{24} - \frac{cos (4 x)}{16} - \frac{cos (2 x)}{8} + C$ $=cos(6x)/24-cos(4x)/16-cos(2x)/8+C$

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$\int sin θ sin 2 θ sin 3 θ$ $int sin theta sin2theta sin3theta$ ?

Evalute:
$\int sin θ sin 2 θ sin 3 θ$ $int sin theta sin2theta sin3theta$

1 Answer

Explanation:

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∫sinθsin2θsin3θint sin theta sin2theta sin3theta ?

Evalute: ∫sinθsin2θsin3θint sin theta sin2theta sin3theta

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