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

The integral of (x+3)^2sin(x) is -x^2cos(x)+2xsin(x)-6xcos(x)-7cos(x)+6sin(x)+C

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

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

- x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x) + C

Solution steps

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

x + 3 = 3 (\frac{x}{3} + 1)

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

Simplify

(3 (\frac{x}{3} + 1))^{2} sin (x) : x^{2} sin (x) + 6 x sin (x) + 9 sin (x)

= \int x^{2} sin (x) + 6 x sin (x) + 9 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 6 x sin (x) d x + \int 9 sin (x) d x

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

\int 6 x sin (x) d x = 6 (- x cos (x) + sin (x))

\int 9 sin (x) d x = - 9 cos (x)

= - x^{2} cos (x) + 2 (x sin (x) + cos (x)) + 6 (- x cos (x) + sin (x)) - 9 cos (x)

Simplify

- x^{2} cos (x) + 2 (x sin (x) + cos (x)) + 6 (- x cos (x) + sin (x)) - 9 cos (x) : - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x)

= - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x)

Add a constant to the solution

= - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x) + C

\frac{d}{d x} (sin (- 4 x))

(x^{2} + 1) \frac{d y}{d x} + 4 x y = x

\frac{( x )}{x + 1}

\int \frac{x}{x ^{4} + 8 x ^{2} + 17} d x

\frac{\partial}{\partial x} (y cos (x))

What is the integral of (x+3)^2sin(x) ?
The integral of (x+3)^2sin(x) is -x^2cos(x)+2xsin(x)-6xcos(x)-7cos(x)+6sin(x)+C