What is the integral of x^3cos(5x) ?

The integral of x^3cos(5x) is 1/5 x^3sin(5x)-3/5 (-1/5 x^2cos(5x)+2/125 (5xsin(5x)+cos(5x)))+C

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integral of x^3cos(5x)

\int x^{3} cos (5 x) d x

\frac{1}{5} x^{3} sin (5 x) - \frac{3}{5} (- \frac{1}{5} x^{2} cos (5 x) + \frac{2}{125} (5 x sin (5 x) + cos (5 x))) + C

Solution steps

\int x^{3} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} x^{3} sin (5 x) - \int \frac{3}{5} x^{2} sin (5 x) d x

= \frac{1}{5} x^{3} sin (5 x) - \int \frac{3}{5} x^{2} sin (5 x) d x

\int \frac{3}{5} x^{2} sin (5 x) d x = \frac{3}{5} (- \frac{1}{5} x^{2} cos (5 x) + \frac{2}{125} (5 x sin (5 x) + cos (5 x)))

= \frac{1}{5} x^{3} sin (5 x) - \frac{3}{5} (- \frac{1}{5} x^{2} cos (5 x) + \frac{2}{125} (5 x sin (5 x) + cos (5 x)))

Add a constant to the solution

= \frac{1}{5} x^{3} sin (5 x) - \frac{3}{5} (- \frac{1}{5} x^{2} cos (5 x) + \frac{2}{125} (5 x sin (5 x) + cos (5 x))) + C

\int 20 x^{9} d x

f (x) = 4 x - 1, at x = 16

y = 3 x, y = x^{2} - 4

y^{^{'}} = x + 6 y

\frac{d}{d x} (ln (\frac{x}{3}) + \frac{x - 3}{x})

What is the integral of x^3cos(5x) ?
The integral of x^3cos(5x) is 1/5 x^3sin(5x)-3/5 (-1/5 x^2cos(5x)+2/125 (5xsin(5x)+cos(5x)))+C