What is the integral of x^4sin(3x) ?

The integral of x^4sin(3x) is -1/3 x^4cos(3x)+4/81 (9x^3sin(3x)+9x^2cos(3x)-6xsin(3x)-2cos(3x))+C

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integral of x^4sin(3x)

\int x^{4} sin (3 x) d x

- \frac{1}{3} x^{4} cos (3 x) + \frac{4}{81} (9 x^{3} sin (3 x) + 9 x^{2} cos (3 x) - 6 x sin (3 x) - 2 cos (3 x)) + C

Solution steps

\int x^{4} sin (3 x) d x

Apply Integration By Parts

: - \frac{1}{3} x^{4} cos (3 x) - \int - \frac{4}{3} x^{3} cos (3 x) d x

= - \frac{1}{3} x^{4} cos (3 x) - \int - \frac{4}{3} x^{3} cos (3 x) d x

\int - \frac{4}{3} x^{3} cos (3 x) d x = - \frac{4}{81} (9 x^{3} sin (3 x) + 9 x^{2} cos (3 x) - 6 x sin (3 x) - 2 cos (3 x))

= - \frac{1}{3} x^{4} cos (3 x) - (- \frac{4}{81} (9 x^{3} sin (3 x) + 9 x^{2} cos (3 x) - 6 x sin (3 x) - 2 cos (3 x)))

Simplify

= - \frac{1}{3} x^{4} cos (3 x) + \frac{4}{81} (9 x^{3} sin (3 x) + 9 x^{2} cos (3 x) - 6 x sin (3 x) - 2 cos (3 x))

Add a constant to the solution

= - \frac{1}{3} x^{4} cos (3 x) + \frac{4}{81} (9 x^{3} sin (3 x) + 9 x^{2} cos (3 x) - 6 x sin (3 x) - 2 cos (3 x)) + C

\int - (x^{2} + 2) e^{x} d x

\frac{\partial}{\partial x} (3^{2 y})

f (x) = 2^{70}

\int \frac{x ^{5} - 3 x ^{2} + 4}{x} d x

\frac{d}{d x} (- \frac{x}{9 - x ^{2}})

What is the integral of x^4sin(3x) ?
The integral of x^4sin(3x) is -1/3 x^4cos(3x)+4/81 (9x^3sin(3x)+9x^2cos(3x)-6xsin(3x)-2cos(3x))+C