What is the integral from 0 to 1 of sin^2(pi/6 x) ?

The integral from 0 to 1 of sin^2(pi/6 x) is (2pi-3sqrt(3))/(4pi)

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integral from 0 to 1 of sin^2(pi/6 x)

\int_{0}^{1} sin^{2} (\frac{π}{6} x) d x

\frac{2 π - 3 3}{4 π}

Decimal

0.08650 \dots

Solution steps

\int_{0}^{1} sin^{2} (\frac{π}{6} x) d x

Rewrite using trig identities

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

Take the constant out:

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

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

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}^{1} 1 d x - \int_{0}^{1} cos (\frac{π}{3} x) d x)

\int_{0}^{1} 1 d x = 1

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

= \frac{1}{2} (1 - \frac{3 3}{2 π})

Simplify

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

= \frac{2 π - 3 3}{4 π}

What is the integral from 0 to 1 of sin^2(pi/6 x) ?
The integral from 0 to 1 of sin^2(pi/6 x) is (2pi-3sqrt(3))/(4pi)