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

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

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integral of 1/4 sin(4x)

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

- \frac{1}{16} cos (4 x) + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

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

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

Add a constant to the solution

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

s l o p e (- 3, 3), (4, - 5)

x \to - 4 lim ((x + 3)^{2})

\int \frac{x}{x ^{2} + x + 1} d x

\frac{d}{d x} (- e^{3 x})

d er i v a t i v e \frac{( 6 x ^{2} + 2 x + 6 )}{x}

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