What is the integral of 1/3 x^3sin(2x) ?

The integral of 1/3 x^3sin(2x) is -1/6 x^3cos(2x)+1/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))+C

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integral of 1/3 x^3sin(2x)

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

- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

Solution steps

\int \frac{1}{3} 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

= \frac{1}{3} \cdot \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}{3} (- \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}{3} (- \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}{3} (- \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}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

Add a constant to the solution

= - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

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What is the integral of 1/3 x^3sin(2x) ?
The integral of 1/3 x^3sin(2x) is -1/6 x^3cos(2x)+1/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))+C