What is the integral of xcos(4x) ?

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

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integral of xcos(4x)

\int x 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

x^{2} + 3 x + 9

n = 1 \sum \infty \frac{x ^{n}}{n + 3}

\frac{1}{2 s ^{3}}

\int_{0}^{5} \frac{1}{x ^{\frac{18}{17}}} d x

\frac{d}{d x} (2^{x y})

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