What is the integral of (4x^2sin(2x))/9 ?

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

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integral of (4x^2sin(2x))/9

\int \frac{4 x ^{2} sin ( 2 x )}{9} d x

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

Solution steps

\int \frac{4 x ^{2} sin ( 2 x )}{9} d x

Take the constant out:

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

= \frac{4}{9} \cdot \int x^{2} sin (2 x) d x

Apply Integration By Parts

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

= \frac{4}{9} (- \frac{1}{2} x^{2} cos (2 x) - \int - x cos (2 x) d x)

\int - x cos (2 x) d x = - \frac{1}{4} (2 x sin (2 x) + cos (2 x))

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

Simplify

\frac{4}{9} (- \frac{1}{2} x^{2} cos (2 x) - (- \frac{1}{4} (2 x sin (2 x) + cos (2 x)))) : - \frac{2}{9} x^{2} cos (2 x) + \frac{1}{9} (2 x sin (2 x) + cos (2 x))

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

Add a constant to the solution

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

n = 2 \sum \infty \frac{n - 2}{n 5 ^{n}}

n = 3 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

x \to \infty lim (2 + \frac{5}{x})

\int \frac{e ^{- x^{2}}}{x} d x

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

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