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

The integral of (x^2-4x)cos(4x) is 1/4 x^2sin(4x)-1/32 sin(4x)+1/8 xcos(4x)-xsin(4x)-1/4 cos(4x)+C

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integral of (x^2-4x)cos(4x)

\int (x^{2} - 4 x) cos (4 x) d x

\frac{1}{4} x^{2} sin (4 x) - \frac{1}{32} sin (4 x) + \frac{1}{8} x cos (4 x) - x sin (4 x) - \frac{1}{4} cos (4 x) + C

Solution steps

\int (x^{2} - 4 x) cos (4 x) d x

Expand

(x^{2} - 4 x) cos (4 x) : x^{2} cos (4 x) - 4 x cos (4 x)

= \int x^{2} cos (4 x) - 4 x cos (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^{2} cos (4 x) d x - \int 4 x cos (4 x) d x

\int x^{2} cos (4 x) d x = \frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

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

= \frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) - (x sin (4 x) + \frac{1}{4} cos (4 x))

Simplify

\frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) - (x sin (4 x) + \frac{1}{4} cos (4 x)) : \frac{1}{4} x^{2} sin (4 x) - \frac{1}{32} sin (4 x) + \frac{1}{8} x cos (4 x) - x sin (4 x) - \frac{1}{4} cos (4 x)

= \frac{1}{4} x^{2} sin (4 x) - \frac{1}{32} sin (4 x) + \frac{1}{8} x cos (4 x) - x sin (4 x) - \frac{1}{4} cos (4 x)

Add a constant to the solution

= \frac{1}{4} x^{2} sin (4 x) - \frac{1}{32} sin (4 x) + \frac{1}{8} x cos (4 x) - x sin (4 x) - \frac{1}{4} cos (4 x) + C

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What is the integral of (x^2-4x)cos(4x) ?
The integral of (x^2-4x)cos(4x) is 1/4 x^2sin(4x)-1/32 sin(4x)+1/8 xcos(4x)-xsin(4x)-1/4 cos(4x)+C