What is the integral of 4cos(6x) ?

The integral of 4cos(6x) is 2/3 sin(6x)+C

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integral of 4cos(6x)

\int 4 cos (6 x) d x

\frac{2}{3} sin (6 x) + C

Solution steps

\int 4 cos (6 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 (6 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{6} d u

= 4 \cdot \int cos (u) \frac{1}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int cos (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{6} sin (u)

Substitute back

u = 6 x

= 4 \cdot \frac{1}{6} sin (6 x)

Simplify

4 \cdot \frac{1}{6} sin (6 x) : \frac{2}{3} sin (6 x)

= \frac{2}{3} sin (6 x)

Add a constant to the solution

= \frac{2}{3} sin (6 x) + C

\int - \frac{3 x}{2} d x

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\int_{0}^{2} 4 - x^{2} d x

\frac{\partial}{\partial y} (x y^{2})

3 ln (2 x)