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

The integral of 2xcos(4x) is 1/8 (4xsin(4x)+cos(4x))+C

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integral of 2xcos(4x)

\int 2 x cos (4 x) d x

\frac{1}{8} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int 2 x cos (4 x) d x

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= 2 \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= 2 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

2 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : \frac{1}{8} (4 x sin (4 x) + cos (4 x))

= \frac{1}{8} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{8} (4 x sin (4 x) + cos (4 x)) + C

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What is the integral of 2xcos(4x) ?
The integral of 2xcos(4x) is 1/8 (4xsin(4x)+cos(4x))+C