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

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

= \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

(- 3, 3), (4, - 5)

\frac{9}{x + 1}

x \to - 1 lim ((x + 2)^{7} (2 x)^{2})

\frac{\partial}{\partial x} (\frac{3}{2 x + 3 y})

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

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