What is the integral of (x^2+6x)cos(2x) ?

The integral of (x^2+6x)cos(2x) is 1/2 x^2sin(2x)+3xsin(2x)+1/2 xcos(2x)+3/2 cos(2x)-1/4 sin(2x)+C

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integral of (x^2+6x)cos(2x)

\int (x^{2} + 6 x) cos (2 x) d x

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

Solution steps

\int (x^{2} + 6 x) cos (2 x) d x

Apply Integration By Parts

: \frac{1}{2} sin (2 x) (x^{2} + 6 x) - \int sin (2 x) (x + 3) d x

= \frac{1}{2} sin (2 x) (x^{2} + 6 x) - \int sin (2 x) (x + 3) d x

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

= \frac{1}{2} sin (2 x) (x^{2} + 6 x) - (- \frac{1}{2} cos (2 x) (x + 3) + \frac{1}{4} sin (2 x))

Simplify

\frac{1}{2} sin (2 x) (x^{2} + 6 x) - (- \frac{1}{2} cos (2 x) (x + 3) + \frac{1}{4} sin (2 x)) : \frac{1}{2} x^{2} sin (2 x) + 3 x sin (2 x) + \frac{1}{2} x cos (2 x) + \frac{3}{2} cos (2 x) - \frac{1}{4} sin (2 x)

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

Add a constant to the solution

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

\int 8 cos (8 x) sin (8 x) d x

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What is the integral of (x^2+6x)cos(2x) ?
The integral of (x^2+6x)cos(2x) is 1/2 x^2sin(2x)+3xsin(2x)+1/2 xcos(2x)+3/2 cos(2x)-1/4 sin(2x)+C