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

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

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integral of (sin(4x))/4

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

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

Solution steps

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

\frac{d}{d x} ((x^{2} - 2 x - 1) (\frac{x + 1}{x + 3}))

\frac{d y}{d t} = - \frac{2 t}{y}

x \to - 2 lim (lo g_{a} (x))

\int \frac{1 - 3 y}{2 y - 3 y ^{2}} d y

\frac{d}{d x} (lo g_{2} (1 - 3 x))

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