What is the integral of 4cos(3x) ?

The integral of 4cos(3x) is 4/3 sin(3x)+C

integral of 4cos(3x)

\int 4 cos (3 x) d x

\frac{4}{3} sin (3 x) + C

Solution steps

\int 4 cos (3 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 (3 x) d x

Apply u-substitution

= 4 \cdot \int cos (u) \frac{1}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} sin (u)

Substitute back

u = 3 x

= 4 \cdot \frac{1}{3} sin (3 x)

Simplify

4 \cdot \frac{1}{3} sin (3 x) : \frac{4}{3} sin (3 x)

= \frac{4}{3} sin (3 x)

Add a constant to the solution

= \frac{4}{3} sin (3 x) + C

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

d er i v a t i v e 8 e^{x}

\frac{d}{d x} (9 e^{x} + \frac{8}{3 x})

\frac{d}{d t} (10 t e^{- 5 t})

x \to 0 lim (\frac{sin ( k t )}{t})