What is the integral of (8x^2-4x)cos(4x) ?

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

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integral of (8x^2-4x)cos(4x)

\int (8 x^{2} - 4 x) cos (4 x) d x

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

Solution steps

\int (8 x^{2} - 4 x) cos (4 x) d x

Expand

(8 x^{2} - 4 x) cos (4 x) : 8 x^{2} cos (4 x) - 4 x cos (4 x)

= \int 8 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 8 x^{2} cos (4 x) d x - \int 4 x cos (4 x) d x

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

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

Add a constant to the solution

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

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

\int (tan (x)) (ln (cos (x))) d x

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

\int - \frac{x}{1 - x ^{2}} d x

\int x (ln (x))^{4} d x

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