What is the integral of 4x^2cos(6x) ?

The integral of 4x^2cos(6x) is 1/27 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C

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integral of 4x^2cos(6x)

\int 4 x^{2} cos (6 x) d x

\frac{1}{27} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

Solution steps

\int 4 x^{2} cos (6 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 (6 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{216} d u

= 4 \cdot \int \frac{u ^{2} cos ( u )}{216} d u

Take the constant out:

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

= 4 \cdot \frac{1}{216} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

: u^{2} sin (u) - \int 2 u sin (u) d u

= 4 \cdot \frac{1}{216} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= 4 \cdot \frac{1}{216} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 6 x

= 4 \cdot \frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x)))

Simplify

4 \cdot \frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x))) : \frac{1}{27} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

= \frac{1}{27} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

Add a constant to the solution

= \frac{1}{27} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

\frac{d}{d x} (\frac{2 x ^{3}}{2 x ^{4} - 5 x ^{3} + x ^{2}})

f (x) = 4 e^{x}

x (x - 10)

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

\int (x + 3)^{3} d x

What is the integral of 4x^2cos(6x) ?
The integral of 4x^2cos(6x) is 1/27 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C