What is the integral of 4x^2cos(mx) ?

The integral of 4x^2cos(mx) is 4(1/m x^2sin(mx)-2/m (-1/m xcos(mx)+1/(m^2)sin(mx)))+C

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integral of 4x^2cos(mx)

\int 4 x^{2} cos (m x) d x

4 (\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 4 x^{2} cos (m x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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

= 4 (\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))

= 4 (\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

= 4 (\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 x - 3 d x

\int (7 - \frac{1}{x ^{6}})^{2} (\frac{1}{x ^{7}}) d x

f (t) = e^{t} - 1

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

f (x) = \frac{ln ( x )}{x ^{4}}

What is the integral of 4x^2cos(mx) ?
The integral of 4x^2cos(mx) is 4(1/m x^2sin(mx)-2/m (-1/m xcos(mx)+1/(m^2)sin(mx)))+C