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

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

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integral of (x+2)^2sin(x)

\int (x + 2)^{2} sin (x) d x

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

Solution steps

\int (x + 2)^{2} sin (x) d x

Expand

(x + 2)^{2} sin (x) : x^{2} sin (x) + 4 x sin (x) + 4 sin (x)

= \int x^{2} sin (x) + 4 x sin (x) + 4 sin (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 (x) d x + \int 4 x sin (x) d x + \int 4 sin (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{3} - x)

\int \frac{1}{5} - \frac{2}{x} d x

\frac{d y}{d x} = x x - 18

- 3 t sin (2 t)

f (x) = 4 x^{2} + 2 x - 3

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