What is the integral of 1/3 cos(3x) ?

The integral of 1/3 cos(3x) is 1/9 sin(3x)+C

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integral of 1/3 cos(3x)

\int \frac{1}{3} cos (3 x) d x

\frac{1}{9} sin (3 x) + C

Solution steps

\int \frac{1}{3} cos (3 x) d x

Take the constant out:

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

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

Apply u-substitution

: \int cos (u) \frac{1}{3} d u

= \frac{1}{3} \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

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} \cdot \frac{1}{3} sin (3 x) : \frac{1}{9} sin (3 x)

= \frac{1}{9} sin (3 x)

Add a constant to the solution

= \frac{1}{9} sin (3 x) + C

(y^{3} - 1) e^{x} d x + 3 y^{2} (e^{x} + 1) d y = 0, y (0) = 0

\frac{\partial}{\partial c} (c^{\frac{1}{2}} + l^{\frac{1}{2}})

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

x \to \infty lim (x + 1)

\frac{\partial}{\partial x} (- k p (x) gx)

What is the integral of 1/3 cos(3x) ?
The integral of 1/3 cos(3x) is 1/9 sin(3x)+C