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

The integral from 0 to pi/6 of cos^6(3x) is (5pi)/(96)

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

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

\frac{5 π}{96}

Decimal Notation

0.16362 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos^{6} (u) \frac{1}{3} d u

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

Apply Integral Reduction

= \frac{1}{3} ([\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} cos^{4} (u) d u)

\int_{0}^{\frac{π}{2}} cos^{4} (u) d u = \frac{3 π}{16}

= \frac{1}{3} ([\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16})

Simplify

\frac{1}{3} ([\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16}) : \frac{1}{3} ([\frac{1}{6} cos^{5} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32})

= \frac{1}{3} ([\frac{1}{6} cos^{5} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32})

Compute the boundaries:

0

= \frac{1}{3} (0 + \frac{5 π}{32})

Simplify

= \frac{5 π}{96}

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What is the integral from 0 to pi/6 of cos^6(3x) ?
The integral from 0 to pi/6 of cos^6(3x) is (5pi)/(96)