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

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

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integral of 3sin(4x)

\int 3 sin (4 x) d x

- \frac{3}{4} cos (4 x) + C

Solution steps

\int 3 sin (4 x) d x

Take the constant out:

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

= 3 \cdot \int sin (4 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

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

= - \frac{3}{4} cos (4 x)

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 3)

\frac{d}{d x} ((2 x - 1)^{5})

\int \frac{1}{tan ( x ) - sin ( x )} d x

4 x \cdot 2^{x}

\frac{d y}{d x} = 20 - \frac{y}{6000 + 20 x}