What is the integral of sin^7(x/9)cos(x/9) ?

The integral of sin^7(x/9)cos(x/9) is 9/8 sin^8(x/9)+C

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integral of sin^7(x/9)cos(x/9)

\int sin^{7} (\frac{x}{9}) cos (\frac{x}{9}) d x

\frac{9}{8} sin^{8} (\frac{x}{9}) + C

Solution steps

\int sin^{7} (\frac{x}{9}) cos (\frac{x}{9}) d x

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{7}{8} u^{\frac{8}{7}}

= \frac{9}{7} \cdot \frac{7}{8} u^{\frac{8}{7}}

Substitute back

u = sin^{7} (\frac{x}{9})

= \frac{9}{7} \cdot \frac{7}{8} (sin^{7} (\frac{x}{9}))^{\frac{8}{7}}

Simplify

\frac{9}{7} \cdot \frac{7}{8} (sin^{7} (\frac{x}{9}))^{\frac{8}{7}} : \frac{9}{8} sin^{8} (\frac{x}{9})

= \frac{9}{8} sin^{8} (\frac{x}{9})

Add a constant to the solution

= \frac{9}{8} sin^{8} (\frac{x}{9}) + C

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What is the integral of sin^7(x/9)cos(x/9) ?
The integral of sin^7(x/9)cos(x/9) is 9/8 sin^8(x/9)+C