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

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

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

\frac{d}{d x} (x e^{- x} - e^{- x})

n = 1 \sum \infty n 3^{- n}

x \to 0 lim (\frac{1}{3 x})

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