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

The integral of x^3sin(-2x) is 1/2 x^3cos(2x)-3/4 x^2sin(2x)-3/4 xcos(2x)+3/8 sin(2x)+C

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integral of x^3sin(-2x)

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

\frac{1}{2} x^{3} cos (2 x) - \frac{3}{4} x^{2} sin (2 x) - \frac{3}{4} x cos (2 x) + \frac{3}{8} sin (2 x) + C

Solution steps

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

Simplify

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: - \frac{1}{2} x^{3} cos (2 x) - \int - \frac{3}{2} x^{2} cos (2 x) d x

= - (- \frac{1}{2} x^{3} cos (2 x) - \int - \frac{3}{2} x^{2} cos (2 x) d x)

\int - \frac{3}{2} x^{2} cos (2 x) d x = - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= - (- \frac{1}{2} x^{3} cos (2 x) - (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))

Simplify

- (- \frac{1}{2} x^{3} cos (2 x) - (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))) : \frac{1}{2} x^{3} cos (2 x) - \frac{3}{4} x^{2} sin (2 x) - \frac{3}{4} x cos (2 x) + \frac{3}{8} sin (2 x)

= \frac{1}{2} x^{3} cos (2 x) - \frac{3}{4} x^{2} sin (2 x) - \frac{3}{4} x cos (2 x) + \frac{3}{8} sin (2 x)

Add a constant to the solution

= \frac{1}{2} x^{3} cos (2 x) - \frac{3}{4} x^{2} sin (2 x) - \frac{3}{4} x cos (2 x) + \frac{3}{8} sin (2 x) + C

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What is the integral of x^3sin(-2x) ?
The integral of x^3sin(-2x) is 1/2 x^3cos(2x)-3/4 x^2sin(2x)-3/4 xcos(2x)+3/8 sin(2x)+C