What is the integral of sin^2(1/3 θ) ?

The integral of sin^2(1/3 θ) is 1/2 (θ-3/2 sin((2θ)/3))+C

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integral of sin^2(1/3 θ)

\int sin^{2} (\frac{1}{3} θ) d θ

\frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3})) + C

Solution steps

\int sin^{2} (\frac{1}{3} θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (\frac{2}{3} θ)) d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 - cos (\frac{2}{3} θ) d θ

Rewrite using trig identities

= \frac{1}{2} \cdot \int 2 sin^{2} (\frac{θ}{3}) d θ

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int sin^{2} (\frac{θ}{3}) d θ

Rewrite using trig identities

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{2} (1 - cos (\frac{2 θ}{3})) d θ

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 - cos (\frac{2 θ}{3}) d θ

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}{2} \cdot 2 \cdot \frac{1}{2} (\int 1 d θ - \int cos (\frac{2 θ}{3}) d θ)

\int 1 d θ = θ

\int cos (\frac{2 θ}{3}) d θ = \frac{3}{2} sin (\frac{2 θ}{3})

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3}))

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3})) : \frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3}))

= \frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3}))

Add a constant to the solution

= \frac{1}{2} (θ - \frac{3}{2} sin (\frac{2 θ}{3})) + C

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What is the integral of sin^2(1/3 θ) ?
The integral of sin^2(1/3 θ) is 1/2 (θ-3/2 sin((2θ)/3))+C