What is the integral of cos^4(3x)sin(3x) ?

The integral of cos^4(3x)sin(3x) is -1/15 cos^5(3x)+C

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integral of cos^4(3x)sin(3x)

\int cos^{4} (3 x) sin (3 x) d x

- \frac{1}{15} cos^{5} (3 x) + C

Solution steps

\int cos^{4} (3 x) sin (3 x) d x

Apply u-substitution

: \int cos^{4} (u) sin (u) \frac{1}{3} d u

= \int cos^{4} (u) sin (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^{4} (u) sin (u) d u

Apply u-substitution

: \int - v^{4} d v

= \frac{1}{3} \cdot \int - v^{4} 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^{4} d v)

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{3} (- \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (- \frac{cos ^{5} ( 3 x )}{5})

Simplify

\frac{1}{3} (- \frac{cos ^{5} ( 3 x )}{5}) : - \frac{1}{15} cos^{5} (3 x)

= - \frac{1}{15} cos^{5} (3 x)

Add a constant to the solution

= - \frac{1}{15} cos^{5} (3 x) + C

t \frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} = t^{2}

y = - \frac{3}{2} x^{2} + 9 x + 9, (- 1, - \frac{3}{2})

\int 7 - x^{2} d x

x \to - 2 lim (\frac{2 x}{2 x + 2})

\frac{d}{d x} ((1 + (1 + (1 + x^{3})^{4})^{5})^{6})

What is the integral of cos^4(3x)sin(3x) ?
The integral of cos^4(3x)sin(3x) is -1/15 cos^5(3x)+C