What is the integral from 0 to pi/6 of cos^8(3x) ?

The integral from 0 to pi/6 of cos^8(3x) is (35pi)/(768)

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integral from 0 to pi/6 of cos^8(3x)

\int_{0}^{\frac{π}{6}} cos^{8} (3 x) d x

\frac{35 π}{768}

Decimal

0.14317 \dots

Solution steps

\int_{0}^{\frac{π}{6}} cos^{8} (3 x) d x

Apply u-substitution

= \int_{0}^{\frac{π}{2}} cos^{8} (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^{8} (u) d u

Apply Integral Reduction

= \frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \int_{0}^{\frac{π}{2}} cos^{6} (u) d u)

\int_{0}^{\frac{π}{2}} cos^{6} (u) d u = \frac{5 π}{32}

= \frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32})

Simplify

\frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32}) : \frac{1}{3} ([\frac{1}{8} cos^{7} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{35 π}{256})

= \frac{1}{3} ([\frac{1}{8} cos^{7} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{35 π}{256})

Compute the boundaries:

0

= \frac{1}{3} (0 + \frac{35 π}{256})

Simplify

= \frac{35 π}{768}

What is the integral from 0 to pi/6 of cos^8(3x) ?
The integral from 0 to pi/6 of cos^8(3x) is (35pi)/(768)