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

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

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integral of-1/16 sin(4x)

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

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

Solution steps

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

Apply u-substitution

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

= - \frac{1}{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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

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

= \frac{1}{64} cos (4 x)

Add a constant to the solution

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

x \to 64 lim (x^{\frac{4}{3}})

y = x^{2} + 5 x, (- 4, - 4)

\frac{d}{d x} (x - \frac{x ^{2}}{800})

\frac{d}{d x} (x^{3} - 6 x + 12)

\frac{d}{d x} (- 216 cos (3 x))

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