What is the integral of 4xcos(2x) ?

The integral of 4xcos(2x) is 2xsin(2x)+cos(2x)+C

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integral of 4xcos(2x)

\int 4 x cos (2 x) d x

2 x sin (2 x) + cos (2 x) + C

Solution steps

\int 4 x cos (2 x) d x

Take the constant out:

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

= 4 \cdot \int x cos (2 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{4} d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 2 x

= 4 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x)))

Simplify

4 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x))) : 2 x sin (2 x) + cos (2 x)

= 2 x sin (2 x) + cos (2 x)

Add a constant to the solution

= 2 x sin (2 x) + cos (2 x) + C

x \to 2 lim (2 x + 4)

\frac{\partial}{\partial x} (x d y)

- e^{x} + 3 x - 3 sin (x)

\frac{d}{d x} ((sin (x^{2}) + 1)^{\frac{1}{y}})

\frac{\partial}{\partial x} (sin (π (6 x - 3 y)))

What is the integral of 4xcos(2x) ?
The integral of 4xcos(2x) is 2xsin(2x)+cos(2x)+C