What is the integral from 0 to pi/6 of (1-cos(3x))sin(3x) ?

The integral from 0 to pi/6 of (1-cos(3x))sin(3x) is 1/6

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integral from 0 to pi/6 of (1-cos(3x))sin(3x)

\int_{0}^{\frac{π}{6}} (1 - cos (3 x)) sin (3 x) d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\int_{0}^{\frac{π}{6}} (1 - cos (3 x)) sin (3 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u}{3} d u

= \int_{0}^{1} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{1}

= \frac{1}{3} [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{3} \cdot \frac{1}{2}

Simplify

= \frac{1}{6}

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