What is the integral of sin^2(ωt+φ) ?

The integral of sin^2(ωt+φ) is φ/(2ω)+t/2-1/(4ω)sin(2(φ+ωt))+C

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integral of sin^2(ωt+φ)

\int sin^{2} (ω t + φ) d t

\frac{φ}{2 ω} + \frac{t}{2} - \frac{1}{4 ω} sin (2 (φ + ω t)) + C

Solution steps

\int sin^{2} (ω t + φ) d t

Apply u-substitution

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

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

= \frac{1}{ω} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = φ + ω t

= \frac{1}{ω} \cdot \frac{1}{2} (φ + ω t - \frac{1}{2} sin (2 (φ + ω t)))

Simplify

\frac{1}{ω} \cdot \frac{1}{2} (φ + ω t - \frac{1}{2} sin (2 (φ + ω t))) : \frac{φ}{2 ω} + \frac{t}{2} - \frac{1}{4 ω} sin (2 (φ + ω t))

= \frac{φ}{2 ω} + \frac{t}{2} - \frac{1}{4 ω} sin (2 (φ + ω t))

Add a constant to the solution

= \frac{φ}{2 ω} + \frac{t}{2} - \frac{1}{4 ω} sin (2 (φ + ω t)) + C

What is the integral of sin^2(ωt+φ) ?
The integral of sin^2(ωt+φ) is φ/(2ω)+t/2-1/(4ω)sin(2(φ+ωt))+C