What is the integral of-1/16 cos(4x) ?

The integral of-1/16 cos(4x) is -1/64 sin(4x)+C

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integral of-1/16 cos(4x)

\int - \frac{1}{16} cos (4 x) d x

- \frac{1}{64} sin (4 x) + C

Solution steps

\int - \frac{1}{16} cos (4 x) d x

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

- \frac{1}{16} \cdot \frac{1}{4} sin (4 x) : - \frac{1}{64} sin (4 x)

= - \frac{1}{64} sin (4 x)

Add a constant to the solution

= - \frac{1}{64} sin (4 x) + C

What is the integral of-1/16 cos(4x) ?
The integral of-1/16 cos(4x) is -1/64 sin(4x)+C