What is the integral of sin^2(3t)cos(3t) ?

The integral of sin^2(3t)cos(3t) is 1/9 sin^3(3t)+C

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integral of sin^2(3t)cos(3t)

\int sin^{2} (3 t) cos (3 t) d t

\frac{1}{9} sin^{3} (3 t) + C

Solution steps

\int sin^{2} (3 t) cos (3 t) d t

Apply u-substitution

: \int sin^{2} (u) cos (u) \frac{1}{3} d u

= \int sin^{2} (u) cos (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{3} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{3} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{3} \cdot \frac{sin ^{3} ( 3 t )}{3}

Simplify

\frac{1}{3} \cdot \frac{sin ^{3} ( 3 t )}{3} : \frac{1}{9} sin^{3} (3 t)

= \frac{1}{9} sin^{3} (3 t)

Add a constant to the solution

= \frac{1}{9} sin^{3} (3 t) + C

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What is the integral of sin^2(3t)cos(3t) ?
The integral of sin^2(3t)cos(3t) is 1/9 sin^3(3t)+C