What is the integral of 8sin(-4x)cos^3(-4x) ?

The integral of 8sin(-4x)cos^3(-4x) is 1/2 cos^4(4x)+C

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integral of 8sin(-4x)cos^3(-4x)

\int 8 sin (- 4 x) cos^{3} (- 4 x) d x

\frac{1}{2} cos^{4} (4 x) + C

Solution steps

\int 8 sin (- 4 x) cos^{3} (- 4 x) d x

Take the constant out:

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

= 8 \cdot \int sin (- 4 x) cos^{3} (- 4 x) d x

Apply u-substitution

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

= 8 \cdot \int - \frac{1}{4} cos^{3} (u) sin (u) d u

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int cos^{3} (u) sin (u) d u)

Apply u-substitution

: \int - v^{3} d v

= 8 (- \frac{1}{4} \cdot \int - v^{3} d v)

Take the constant out:

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

= 8 (- \frac{1}{4} (- \int v^{3} d v))

Apply the Power Rule

: \frac{v ^{4}}{4}

= 8 (- \frac{1}{4} (- \frac{v ^{4}}{4}))

Substitute back

= 8 (- \frac{1}{4} (- \frac{cos ^{4} ( - 4 x )}{4}))

Simplify

8 (- \frac{1}{4} (- \frac{cos ^{4} ( - 4 x )}{4})) : \frac{1}{2} cos^{4} (4 x)

= \frac{1}{2} cos^{4} (4 x)

Add a constant to the solution

= \frac{1}{2} cos^{4} (4 x) + C

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What is the integral of 8sin(-4x)cos^3(-4x) ?
The integral of 8sin(-4x)cos^3(-4x) is 1/2 cos^4(4x)+C