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

The integral of x*cos(4x) is 1/16 (4xsin(4x)+cos(4x))+C

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integral of x*cos(4x)

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

\frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int x cos (4 x) d x

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 4 x

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

Simplify

= \frac{1}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

y^{^{'}} = y

\frac{4 - x ^{2}}{x ^{3}}

\frac{d}{d x} (sin (x) - x sin (y))

x \to 4 lim (\frac{2 x + 1}{3 x + 3})

\frac{d y}{d x} = \frac{e ^{x}}{y}

What is the integral of x*cos(4x) ?
The integral of x*cos(4x) is 1/16 (4xsin(4x)+cos(4x))+C