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

The integral from 0 to pi/6 of sin^3(6x) is 2/9

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integral from 0 to pi/6 of sin^3(6x)

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

\frac{2}{9}

Solution steps

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

Apply u-substitution

: \int_{0}^{π} sin^{3} (u) \frac{1}{6} d u

= \int_{0}^{π} sin^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{π} sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int_{0}^{π} (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

: \int_{1}^{- 1} - 1 + v^{2} d v

= \frac{1}{6} \cdot \int_{1}^{- 1} - 1 + v^{2} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{6} (- \int_{- 1}^{1} - 1 + v^{2} d v)

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}{6} (- (- \int_{- 1}^{1} 1 d v + \int_{- 1}^{1} v^{2} d v))

\int_{- 1}^{1} 1 d v = 2

\int_{- 1}^{1} v^{2} d v = \frac{2}{3}

= \frac{1}{6} (- (- 2 + \frac{2}{3}))

Simplify

\frac{1}{6} (- (- 2 + \frac{2}{3})) : \frac{2}{9}

= \frac{2}{9}

\int_{0}^{2} \frac{1}{2} x^{3} d x

\int_{1}^{5} 2 π y (5 - y) d y

\int_{- 3}^{3} (2 - x)^{5} d x

\int_{1}^{7} \frac{6}{x ^{4}} d x

\int_{0}^{π} 2 \cdot sin (x) d x

What is the integral from 0 to pi/6 of sin^3(6x) ?
The integral from 0 to pi/6 of sin^3(6x) is 2/9