What is the integral of (cos^4(2x))(-2sin(2x)) ?

The integral of (cos^4(2x))(-2sin(2x)) is 1/5 cos^5(2x)+C

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integral of (cos^4(2x))(-2sin(2x))

\int (cos^{4} (2 x)) (- 2 sin (2 x)) d x

\frac{1}{5} cos^{5} (2 x) + C

Solution steps

\int cos^{4} (2 x) (- 2 sin (2 x)) d x

Rewrite using trig identities

= \int - 4 cos^{4} (2 x) cos (x) sin (x) d x

Take the constant out:

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

= - 4 \cdot \int cos^{4} (2 x) cos (x) sin (x) d x

Rewrite using trig identities

= - 4 \cdot \int \frac{1}{2} cos^{4} (2 x) sin (2 x) d x

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \int cos^{4} (2 x) sin (2 x) d x

Apply u-substitution

: \int cos^{4} (u) sin (u) \frac{1}{2} d u

= - 4 \cdot \frac{1}{2} \cdot \int cos^{4} (u) sin (u) \frac{1}{2} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int cos^{4} (u) sin (u) d u

Apply u-substitution

: \int - v^{4} d v

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{2} \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}{2} \cdot \frac{1}{2} (- \int v^{4} d v)

Apply the Power Rule

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

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{2} (- \frac{v ^{5}}{5})

Substitute back

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5})

Simplify

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

= \frac{1}{5} cos^{5} (2 x)

Add a constant to the solution

= \frac{1}{5} cos^{5} (2 x) + C

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What is the integral of (cos^4(2x))(-2sin(2x)) ?
The integral of (cos^4(2x))(-2sin(2x)) is 1/5 cos^5(2x)+C