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

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

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

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

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - 1 + v^{2} d v

= \frac{1}{3} \cdot \int_{1}^{0} - 1 + v^{2} d v

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

= \frac{1}{3} (- \int_{0}^{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}{3} (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v))

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

\int_{0}^{1} v^{2} d v = \frac{1}{3}

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

Simplify

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

= \frac{2}{9}

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\frac{d y}{d x} = y (1 - y), y (0) = 2

\frac{d}{d x} (\frac{- 2 x ^{2} + 2}{( 1 + x ^{2} ) ^{2}})

\frac{\partial}{\partial x} (- 4 y (- 2 x - y - 3))

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