What is the integral of 3x^2cos(2x) ?

The integral of 3x^2cos(2x) is 3(1/2 x^2sin(2x)+1/2 xcos(2x)-1/4 sin(2x))+C

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integral of 3x^2cos(2x)

\int 3 x^{2} cos (2 x) d x

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

Solution steps

\int 3 x^{2} cos (2 x) d x

Take the constant out:

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

= 3 \cdot \int x^{2} cos (2 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} - 2 x y = 4 x

\int 4 t^{3} - 8 t d t

\int_{1}^{2} x^{3} d x

\frac{d}{d x} (\frac{x - 9}{x + 9})

(\frac{1}{2}, 1), (3 \frac{3}{4}, 3 \frac{1}{2})

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