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

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

Solution steps

\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))

Add a constant to the solution

= - \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)) + C

\frac{d}{d x} (- 3 sin (2 x) - 6 cos (x))

\frac{d}{d x} (sin (25 x^{2} + 6 x + 4))

\frac{d}{d x} (\frac{8 x}{x ^{2} + 4})

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

\frac{d}{d x} (\frac{f ( x )}{g ( x ) + 1})

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