What is the integral of (x-2)sin(4x) ?

The integral of (x-2)sin(4x) is -1/4 xcos(4x)+1/16 sin(4x)+1/2 cos(4x)+C

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integral of (x-2)sin(4x)

\int (x - 2) sin (4 x) d x

- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x) + \frac{1}{2} cos (4 x) + C

Solution steps

\int (x - 2) sin (4 x) d x

Expand

(x - 2) sin (4 x) : x sin (4 x) - 2 sin (4 x)

= \int x sin (4 x) - 2 sin (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x sin (4 x) d x - \int 2 sin (4 x) d x

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

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

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

Simplify

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

Add a constant to the solution

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

What is the integral of (x-2)sin(4x) ?
The integral of (x-2)sin(4x) is -1/4 xcos(4x)+1/16 sin(4x)+1/2 cos(4x)+C