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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int cos (u) sin^{2} (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 cos (u) sin^{2} (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 x )}{3}

Simplify

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

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

Add a constant to the solution

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

f (t) = \frac{4 t}{4 + t}

\int \frac{9}{1 - x ^{2}} d x

\frac{d}{d x} (ln (5 x - 4))

\int \frac{1}{x ^{2} - 6 x + 34} d x

y^{^{''}} + y = - sin (3 x)

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