What is the integral from 0 to pi/6 of cos^7(2x) ?

The integral from 0 to pi/6 of cos^7(2x) is (1181sqrt(3))/(8960)

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integral from 0 to pi/6 of cos^7(2x)

\int_{0}^{\frac{π}{6}} cos^{7} (2 x) d x

\frac{1181 3}{8960}

Decimal Notation

0.22829 \dots

Solution steps

\int_{0}^{\frac{π}{6}} cos^{7} (2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int_{0}^{\frac{3}{2}} 3 v^{2} d v = \frac{3 3}{8}

\int_{0}^{\frac{3}{2}} 3 v^{4} d v = \frac{27 3}{160}

\int_{0}^{\frac{3}{2}} v^{6} d v = \frac{27 3}{896}

= \frac{1}{2} (\frac{3}{2} - \frac{3 3}{8} + \frac{27 3}{160} - \frac{27 3}{896})

Simplify

\frac{1}{2} (\frac{3}{2} - \frac{3 3}{8} + \frac{27 3}{160} - \frac{27 3}{896}) : \frac{1181 3}{8960}

= \frac{1181 3}{8960}

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What is the integral from 0 to pi/6 of cos^7(2x) ?
The integral from 0 to pi/6 of cos^7(2x) is (1181sqrt(3))/(8960)