What is the integral of-4sin(6x)cos^4(6x) ?

The integral of-4sin(6x)cos^4(6x) is 2/15 cos^5(6x)+C

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integral of-4sin(6x)cos^4(6x)

\int - 4 sin (6 x) cos^{4} (6 x) d x

\frac{2}{15} cos^{5} (6 x) + C

Solution steps

\int - 4 sin (6 x) cos^{4} (6 x) d x

Take the constant out:

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

= - 4 \cdot \int sin (6 x) cos^{4} (6 x) d x

Apply u-substitution

: \int sin (u) cos^{4} (u) \frac{1}{6} d u

= - 4 \cdot \int sin (u) cos^{4} (u) \frac{1}{6} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{6} \cdot \int sin (u) cos^{4} (u) d u

Apply u-substitution

: \int - v^{4} d v

= - 4 \cdot \frac{1}{6} \cdot \int - v^{4} d v

Take the constant out:

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

= - 4 \cdot \frac{1}{6} (- \int v^{4} d v)

Apply the Power Rule

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

= - 4 \cdot \frac{1}{6} (- \frac{v ^{5}}{5})

Substitute back

= - 4 \cdot \frac{1}{6} (- \frac{cos ^{5} ( 6 x )}{5})

Simplify

- 4 \cdot \frac{1}{6} (- \frac{cos ^{5} ( 6 x )}{5}) : \frac{2}{15} cos^{5} (6 x)

= \frac{2}{15} cos^{5} (6 x)

Add a constant to the solution

= \frac{2}{15} cos^{5} (6 x) + C

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What is the integral of-4sin(6x)cos^4(6x) ?
The integral of-4sin(6x)cos^4(6x) is 2/15 cos^5(6x)+C