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

The integral of sin^6(x/8)cos(x/8) is 8/7 sin^7(x/8)+C

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

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

\frac{8}{7} sin^{7} (\frac{x}{8}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int v^{6} d v

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

Apply the Power Rule

: \frac{v ^{7}}{7}

= 8 \cdot \frac{v ^{7}}{7}

Substitute back

= 8 \cdot \frac{sin ^{7} ( \frac{x}{8} )}{7}

Simplify

8 \cdot \frac{sin ^{7} ( \frac{x}{8} )}{7} : \frac{8}{7} sin^{7} (\frac{x}{8})

= \frac{8}{7} sin^{7} (\frac{x}{8})

Add a constant to the solution

= \frac{8}{7} sin^{7} (\frac{x}{8}) + C

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What is the integral of sin^6(x/8)cos(x/8) ?
The integral of sin^6(x/8)cos(x/8) is 8/7 sin^7(x/8)+C