What is the integral of sin^8(x/6)cos(x/6) ?

The integral of sin^8(x/6)cos(x/6) is 2/3 sin^9(x/6)+C

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integral of sin^8(x/6)cos(x/6)

\int sin^{8} (\frac{x}{6}) cos (\frac{x}{6}) d x

\frac{2}{3} sin^{9} (\frac{x}{6}) + C

Solution steps

\int sin^{8} (\frac{x}{6}) cos (\frac{x}{6}) d x

Apply u-substitution

: \int sin^{8} (u) cos (u) \cdot 6 d u

= \int sin^{8} (u) cos (u) \cdot 6 d u

Take the constant out:

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

= 6 \cdot \int sin^{8} (u) cos (u) d u

Apply u-substitution

: \int v^{8} d v

= 6 \cdot \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= 6 \cdot \frac{v ^{9}}{9}

Substitute back

= 6 \cdot \frac{sin ^{9} ( \frac{x}{6} )}{9}

Simplify

6 \cdot \frac{sin ^{9} ( \frac{x}{6} )}{9} : \frac{2}{3} sin^{9} (\frac{x}{6})

= \frac{2}{3} sin^{9} (\frac{x}{6})

Add a constant to the solution

= \frac{2}{3} sin^{9} (\frac{x}{6}) + C

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What is the integral of sin^8(x/6)cos(x/6) ?
The integral of sin^8(x/6)cos(x/6) is 2/3 sin^9(x/6)+C