What is the integral of (cos(4x))/8 ?

The integral of (cos(4x))/8 is 1/32 sin(4x)+C

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integral of (cos(4x))/8

\int \frac{cos ( 4 x )}{8} d x

\frac{1}{32} sin (4 x) + C

Solution steps

\int \frac{cos ( 4 x )}{8} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int cos (4 x) d x

Apply u-substitution

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

= \frac{1}{8} \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}{8} \cdot \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{8} \cdot \frac{1}{4} sin (u)

Substitute back

u = 4 x

= \frac{1}{8} \cdot \frac{1}{4} sin (4 x)

Simplify

\frac{1}{8} \cdot \frac{1}{4} sin (4 x) : \frac{1}{32} sin (4 x)

= \frac{1}{32} sin (4 x)

Add a constant to the solution

= \frac{1}{32} sin (4 x) + C

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What is the integral of (cos(4x))/8 ?
The integral of (cos(4x))/8 is 1/32 sin(4x)+C