What is the integral from 0 to pi/(12) of 5cos^5(6x) ?

The integral from 0 to pi/(12) of 5cos^5(6x) is 4/9

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integral from 0 to pi/(12) of 5cos^5(6x)

\int_{0}^{\frac{π}{12}} 5 cos^{5} (6 x) d x

\frac{4}{9}

Decimal Notation

0.44444 \dots

Solution steps

\int_{0}^{\frac{π}{12}} 5 cos^{5} (6 x) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{12}} cos^{5} (6 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos^{5} (u) \frac{1}{6} d u

= 5 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (u) \frac{1}{6} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} cos^{5} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

= 5 \cdot \frac{1}{6} \cdot \int_{0}^{1} (1 - v^{2})^{2} d v

Expand

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 5 \cdot \frac{1}{6} \cdot \int_{0}^{1} 1 - 2 v^{2} + v^{4} 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

= 5 \cdot \frac{1}{6} (\int_{0}^{1} 1 d v - \int_{0}^{1} 2 v^{2} d v + \int_{0}^{1} v^{4} d v)

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

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

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= 5 \cdot \frac{1}{6} (1 - \frac{2}{3} + \frac{1}{5})

Simplify

5 \cdot \frac{1}{6} (1 - \frac{2}{3} + \frac{1}{5}) : \frac{4}{9}

= \frac{4}{9}

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What is the integral from 0 to pi/(12) of 5cos^5(6x) ?
The integral from 0 to pi/(12) of 5cos^5(6x) is 4/9