What is the integral of (x^2-2x-3)sin(2x) ?

The integral of (x^2-2x-3)sin(2x) is -1/2 x^2cos(2x)+1/2 xsin(2x)+xcos(2x)+7/4 cos(2x)-1/2 sin(2x)+C

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integral of (x^2-2x-3)sin(2x)

\int (x^{2} - 2 x - 3) sin (2 x) d x

- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) + x cos (2 x) + \frac{7}{4} cos (2 x) - \frac{1}{2} sin (2 x) + C

Solution steps

\int (x^{2} - 2 x - 3) sin (2 x) d x

Expand

(x^{2} - 2 x - 3) sin (2 x) : x^{2} sin (2 x) - 2 x sin (2 x) - 3 sin (2 x)

= \int x^{2} sin (2 x) - 2 x sin (2 x) - 3 sin (2 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} sin (2 x) d x - \int 2 x sin (2 x) d x - \int 3 sin (2 x) d x

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

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

\int 3 sin (2 x) d x = - \frac{3}{2} cos (2 x)

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)) - (- x cos (2 x) + \frac{1}{2} sin (2 x)) - (- \frac{3}{2} cos (2 x))

Simplify

- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)) - (- x cos (2 x) + \frac{1}{2} sin (2 x)) - (- \frac{3}{2} cos (2 x)) : - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) + x cos (2 x) + \frac{7}{4} cos (2 x) - \frac{1}{2} sin (2 x)

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) + x cos (2 x) + \frac{7}{4} cos (2 x) - \frac{1}{2} sin (2 x)

Add a constant to the solution

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) + x cos (2 x) + \frac{7}{4} cos (2 x) - \frac{1}{2} sin (2 x) + C

What is the integral of (x^2-2x-3)sin(2x) ?
The integral of (x^2-2x-3)sin(2x) is -1/2 x^2cos(2x)+1/2 xsin(2x)+xcos(2x)+7/4 cos(2x)-1/2 sin(2x)+C