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

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

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

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

0

Solution steps

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

Apply u-substitution

: \int_{0}^{π} cos^{3} (u) \frac{1}{6} d u

= \int_{0}^{π} cos^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{π} cos^{3} (u) d u

Simplify

cos^{3} (u) : cos^{2} (u) cos (u)

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

Rewrite using trig identities

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

Apply Integration By Parts

: [- \int - sin (2 u) sin (u) d u + cos^{2} (u) sin (u)]_{0}^{π}

= \frac{1}{6} [- \int - sin (2 u) sin (u) d u + cos^{2} (u) sin (u)]_{0}^{π}

\int - sin (2 u) sin (u) d u = - \frac{2}{3} sin^{3} (u)

= \frac{1}{6} [- (- \frac{2}{3} sin^{3} (u)) + cos^{2} (u) sin (u)]_{0}^{π}

Simplify

\frac{1}{6} [- (- \frac{2}{3} sin^{3} (u)) + cos^{2} (u) sin (u)]_{0}^{π} : \frac{1}{6} [\frac{1}{3} (2 sin^{3} (u) + 3 cos^{2} (u) sin (u))]_{0}^{π}

= \frac{1}{6} [\frac{1}{3} (2 sin^{3} (u) + 3 cos^{2} (u) sin (u))]_{0}^{π}

Compute the boundaries:

0

= \frac{1}{6} \cdot 0

Simplify

= 0

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