What is the integral of sin^3(9x)cos(9x) ?

The integral of sin^3(9x)cos(9x) is 1/36 sin^4(9x)+C

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integral of sin^3(9x)cos(9x)

\int sin^{3} (9 x) cos (9 x) d x

\frac{1}{36} sin^{4} (9 x) + C

Solution steps

\int sin^{3} (9 x) cos (9 x) d x

Apply u-substitution

: \int \frac{3 u}{27} d u

= \int \frac{3 u}{27} d u

Take the constant out:

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

= \frac{1}{27} \cdot \int 3 u d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{27} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = sin^{3} (9 x)

= \frac{1}{27} \cdot \frac{3}{4} (sin^{3} (9 x))^{\frac{4}{3}}

Simplify

\frac{1}{27} \cdot \frac{3}{4} (sin^{3} (9 x))^{\frac{4}{3}} : \frac{1}{36} sin^{4} (9 x)

= \frac{1}{36} sin^{4} (9 x)

Add a constant to the solution

= \frac{1}{36} sin^{4} (9 x) + C

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What is the integral of sin^3(9x)cos(9x) ?
The integral of sin^3(9x)cos(9x) is 1/36 sin^4(9x)+C