What is the integral from 0 to 2pi of sin^2(1/3 θ) ?

The integral from 0 to 2pi of sin^2(1/3 θ) is (8pi+3sqrt(3))/8

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integral from 0 to 2pi of sin^2(1/3 θ)

\int_{0}^{2 π} sin^{2} (\frac{1}{3} θ) d θ

\frac{8 π + 3 3}{8}

Decimal Notation

3.79111 \dots

Solution steps

\int_{0}^{2 π} sin^{2} (\frac{1}{3} θ) d θ

Rewrite using trig identities

= \int_{0}^{2 π} \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_{0}^{2 π} 1 - cos (\frac{2}{3} θ) 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} (\int_{0}^{2 π} 1 d θ - \int_{0}^{2 π} cos (\frac{2}{3} θ) d θ)

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} cos (\frac{2}{3} θ) d θ = - \frac{3 3}{4}

= \frac{1}{2} (2 π - (- \frac{3 3}{4}))

Simplify

\frac{1}{2} (2 π - (- \frac{3 3}{4})) : \frac{8 π + 3 3}{8}

= \frac{8 π + 3 3}{8}

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What is the integral from 0 to 2pi of sin^2(1/3 θ) ?
The integral from 0 to 2pi of sin^2(1/3 θ) is (8pi+3sqrt(3))/8