What is the integral of sin^4(x/9)cos(x/9) ?

The integral of sin^4(x/9)cos(x/9) is 9/5 sin^5(x/9)+C

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integral of sin^4(x/9)cos(x/9)

\int sin^{4} (\frac{x}{9}) cos (\frac{x}{9}) d x

\frac{9}{5} sin^{5} (\frac{x}{9}) + C

Solution steps

\int sin^{4} (\frac{x}{9}) cos (\frac{x}{9}) d x

Apply u-substitution

: \int sin^{4} (u) cos (u) \cdot 9 d u

= \int sin^{4} (u) cos (u) \cdot 9 d u

Take the constant out:

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

= 9 \cdot \int sin^{4} (u) cos (u) d u

Apply u-substitution

: \int v^{4} d v

= 9 \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= 9 \cdot \frac{v ^{5}}{5}

Substitute back

= 9 \cdot \frac{sin ^{5} ( \frac{x}{9} )}{5}

Simplify

9 \cdot \frac{sin ^{5} ( \frac{x}{9} )}{5} : \frac{9}{5} sin^{5} (\frac{x}{9})

= \frac{9}{5} sin^{5} (\frac{x}{9})

Add a constant to the solution

= \frac{9}{5} sin^{5} (\frac{x}{9}) + C

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What is the integral of sin^4(x/9)cos(x/9) ?
The integral of sin^4(x/9)cos(x/9) is 9/5 sin^5(x/9)+C