What is the integral of (4x-1)sin(4x) ?

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

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integral of (4x-1)sin(4x)

\int (4 x - 1) sin (4 x) d x

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

Solution steps

\int (4 x - 1) sin (4 x) d x

Expand

(4 x - 1) sin (4 x) : 4 x sin (4 x) - sin (4 x)

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

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

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

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

Simplify

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

Add a constant to the solution

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

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