What is the integral of x^3sin(6x) ?

The integral of x^3sin(6x) is -1/6 x^3cos(6x)+1/216 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C

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integral of x^3sin(6x)

\int x^{3} sin (6 x) d x

- \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

Solution steps

\int x^{3} sin (6 x) d x

Apply Integration By Parts

: - \frac{1}{6} x^{3} cos (6 x) - \int - \frac{1}{2} x^{2} cos (6 x) d x

= - \frac{1}{6} x^{3} cos (6 x) - \int - \frac{1}{2} x^{2} cos (6 x) d x

\int - \frac{1}{2} x^{2} cos (6 x) d x = - \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

= - \frac{1}{6} x^{3} cos (6 x) - (- \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)))

Simplify

= - \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

Add a constant to the solution

= - \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

f (x) = x^{2} - \frac{1}{x}

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

\frac{\partial}{\partial x} (\frac{4 x - 4 z}{3 y + 4 z})

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

\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}

What is the integral of x^3sin(6x) ?
The integral of x^3sin(6x) is -1/6 x^3cos(6x)+1/216 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C