What is the integral of cos^2(ωt+φ)sin(ωt+φ) ?

The integral of cos^2(ωt+φ)sin(ωt+φ) is -1/(3ω)cos^3(φ+ωt)+C

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integral of cos^2(ωt+φ)sin(ωt+φ)

\int cos^{2} (ω t + φ) sin (ω t + φ) d t

- \frac{1}{3 ω} cos^{3} (φ + ω t) + C

Solution steps

\int cos^{2} (ω t + φ) sin (ω t + φ) d t

Apply u-substitution

: \int \frac{cos ^{2} ( u ) sin ( u )}{ω} d u

= \int \frac{cos ^{2} ( u ) sin ( u )}{ω} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{ω} \cdot \int cos^{2} (u) sin (u) d u

Apply u-substitution

: \int - v^{2} d v

= \frac{1}{ω} \cdot \int - v^{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{ω} (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{ω} (- \frac{v ^{3}}{3})

Substitute back

= \frac{1}{ω} (- \frac{cos ^{3} ( φ + ω t )}{3})

Simplify

\frac{1}{ω} (- \frac{cos ^{3} ( φ + ω t )}{3}) : - \frac{1}{3 ω} cos^{3} (φ + ω t)

= - \frac{1}{3 ω} cos^{3} (φ + ω t)

Add a constant to the solution

= - \frac{1}{3 ω} cos^{3} (φ + ω t) + C

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What is the integral of cos^2(ωt+φ)sin(ωt+φ) ?
The integral of cos^2(ωt+φ)sin(ωt+φ) is -1/(3ω)cos^3(φ+ωt)+C