What is the integral from 0 to pi/6 of cos^7(3x) ?

The integral from 0 to pi/6 of cos^7(3x) is 16/105

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integral from 0 to pi/6 of cos^7(3x)

\int_{0}^{\frac{π}{6}} cos^{7} (3 x) d x

\frac{16}{105}

Decimal

0.15238 \dots

Solution steps

\int_{0}^{\frac{π}{6}} cos^{7} (3 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos^{7} (u) \frac{1}{3} d u

= \int_{0}^{\frac{π}{2}} cos^{7} (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{π}{2}} cos^{7} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int_{0}^{1} (1 - v^{2})^{3} d v

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

Expand

(1 - v^{2})^{3} : 1 - 3 v^{2} + 3 v^{4} - v^{6}

= \frac{1}{3} \cdot \int_{0}^{1} 1 - 3 v^{2} + 3 v^{4} - v^{6} 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} 3 v^{2} d v + \int_{0}^{1} 3 v^{4} d v - \int_{0}^{1} v^{6} d v)

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

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

\int_{0}^{1} 3 v^{4} d v = \frac{3}{5}

\int_{0}^{1} v^{6} d v = \frac{1}{7}

= \frac{1}{3} (1 - 1 + \frac{3}{5} - \frac{1}{7})

Simplify

\frac{1}{3} (1 - 1 + \frac{3}{5} - \frac{1}{7}) : \frac{16}{105}

= \frac{16}{105}

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What is the integral from 0 to pi/6 of cos^7(3x) ?
The integral from 0 to pi/6 of cos^7(3x) is 16/105