What is the integral of (-16sin(4x)) ?

The integral of (-16sin(4x)) is 4cos(4x)+C

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integral of (-16sin(4x))

\int (- 16 sin (4 x)) d x

4 cos (4 x) + C

Solution steps

\int - 16 sin (4 x) d x

Take the constant out:

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

= - 16 \cdot \int sin (4 x) d x

Apply u-substitution

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

= - 16 \cdot \int sin (u) \frac{1}{4} d u

Take the constant out:

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

= - 16 \cdot \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= - 16 \cdot \frac{1}{4} (- cos (u))

Substitute back

u = 4 x

= - 16 \cdot \frac{1}{4} (- cos (4 x))

Simplify

- 16 \cdot \frac{1}{4} (- cos (4 x)) : 4 cos (4 x)

= 4 cos (4 x)

Add a constant to the solution

= 4 cos (4 x) + C

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What is the integral of (-16sin(4x)) ?
The integral of (-16sin(4x)) is 4cos(4x)+C