What is the integral of x^2cos(mx) ?

The integral of x^2cos(mx) is 1/m x^2sin(mx)-2/m (-1/m xcos(mx)+1/(m^2)sin(mx))+C

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integral of x^2cos(mx)

\int x^{2} cos (m x) d x

\frac{1}{m} x^{2} sin (m x) - \frac{2}{m} (- \frac{1}{m} x cos (m x) + \frac{1}{m ^{2}} sin (m x)) + C

Solution steps

\int x^{2} cos (m x) d x

Apply Integration By Parts

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

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

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

= \frac{1}{m} x^{2} sin (m x) - \frac{2}{m} (- \frac{1}{m} x cos (m x) + \frac{1}{m ^{2}} sin (m x))

Add a constant to the solution

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

\int \frac{x + 4}{x} d x

\frac{12}{s ( s ^{2} - 8 s + 12 )}

\int \frac{5}{x ( x ^{4} + 9 )} d x

f (x) = sin (x), at x = \frac{π}{4}

\frac{d}{d x} (arcsin (\frac{1}{3} x))

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