What is the integral of 4cos(9x) ?

The integral of 4cos(9x) is 4/9 sin(9x)+C

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integral of 4cos(9x)

\int 4 cos (9 x) d x

\frac{4}{9} sin (9 x) + C

Solution steps

\int 4 cos (9 x) d x

Take the constant out:

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

= 4 \cdot \int cos (9 x) d x

Apply u-substitution

= 4 \cdot \int cos (u) \frac{1}{9} d u

Take the constant out:

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

= 4 \cdot \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{9} sin (u)

Substitute back

u = 9 x

= 4 \cdot \frac{1}{9} sin (9 x)

Simplify

4 \cdot \frac{1}{9} sin (9 x) : \frac{4}{9} sin (9 x)

= \frac{4}{9} sin (9 x)

Add a constant to the solution

= \frac{4}{9} sin (9 x) + C