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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - v^{2} d v

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

Take the constant out:

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

= \frac{1}{3} (- \int v^{2} d v)

Apply the Power Rule

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

= \frac{1}{3} (- \frac{v ^{3}}{3})

Substitute back

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3})

Simplify

\frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3}) : - \frac{1}{9} cos^{3} (3 x)

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

Add a constant to the solution

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

\int_{- 1}^{4} (5 x^{4} - 8 x^{3} + 6) d x

f^{^{'}} (x) = e^{x}

\int - 6 x e^{x} d x

\int \frac{e ^{x} ( e ^{x} + 1 )}{( e ^{x} - 1 ) ^{3}} d x

\frac{d y}{d x} = 3 (2 - y)

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