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

The integral of 3x(cos(4x)) is 3/16 (4xsin(4x)+cos(4x))+C

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integral of 3x(cos(4x))

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

\frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{u cos ( u )}{16} d u

= 3 \cdot \int \frac{u cos ( u )}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= 3 \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= 3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : \frac{3}{16} (4 x sin (4 x) + cos (4 x))

= \frac{3}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

\int cos (x^{4} + 2) d x

\frac{d y}{d x} = \frac{y}{10} (500 - y)

\frac{\partial}{\partial x} (- sin (x^{2} y) \cdot 2 x y)

\int \frac{x}{1 + 4 x ^{2}} d x

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

What is the integral of 3x(cos(4x)) ?
The integral of 3x(cos(4x)) is 3/16 (4xsin(4x)+cos(4x))+C