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

The integral from 0 to (3pi)/2 of cos^3(9x) is -2/27

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integral from 0 to (3pi)/2 of cos^3(9x)

\int_{0}^{\frac{3 π}{2}} cos^{3} (9 x) d x

- \frac{2}{27}

Decimal Notation

- 0.07407 \dots

Solution steps

\int_{0}^{\frac{3 π}{2}} cos^{3} (9 x) d x

Apply u-substitution

: \int_{0}^{\frac{27 π}{2}} cos^{3} (u) \frac{1}{9} d u

= \int_{0}^{\frac{27 π}{2}} cos^{3} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{\frac{27 π}{2}} cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int_{0}^{\frac{27 π}{2}} (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= \frac{1}{9} \cdot \int_{0}^{- 1} 1 - v^{2} d v

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

= \frac{1}{9} (- \int_{- 1}^{0} 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}{9} (- (\int_{- 1}^{0} 1 d v - \int_{- 1}^{0} v^{2} d v))

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

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

= \frac{1}{9} (- (1 - \frac{1}{3}))

Simplify

\frac{1}{9} (- (1 - \frac{1}{3})) : - \frac{2}{27}

= - \frac{2}{27}

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What is the integral from 0 to (3pi)/2 of cos^3(9x) ?
The integral from 0 to (3pi)/2 of cos^3(9x) is -2/27