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

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

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integral of-16sin(-4t)

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

- 4 cos (4 t) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 4 t

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

Simplify

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

= - 4 cos (4 t)

Add a constant to the solution

= - 4 cos (4 t) + C

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