What is the integral from 0 to pi/(12) of sin^2(3x) ?

The integral from 0 to pi/(12) of sin^2(3x) is (pi-2)/(24)

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integral from 0 to pi/(12) of sin^2(3x)

\int_{0}^{\frac{π}{12}} sin^{2} (3 x) d x

\frac{π - 2}{24}

Decimal

0.04756 \dots

Solution steps

\int_{0}^{\frac{π}{12}} sin^{2} (3 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{12}} \frac{1}{2} (1 - cos (6 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}^{\frac{π}{12}} 1 - cos (6 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}^{\frac{π}{12}} 1 d x - \int_{0}^{\frac{π}{12}} cos (6 x) d x)

\int_{0}^{\frac{π}{12}} 1 d x = \frac{π}{12}

\int_{0}^{\frac{π}{12}} cos (6 x) d x = \frac{1}{6}

= \frac{1}{2} (\frac{π}{12} - \frac{1}{6})

Simplify

\frac{1}{2} (\frac{π}{12} - \frac{1}{6}) : \frac{π - 2}{24}

= \frac{π - 2}{24}

What is the integral from 0 to pi/(12) of sin^2(3x) ?
The integral from 0 to pi/(12) of sin^2(3x) is (pi-2)/(24)