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

The integral of 2x^2cos(3x) is 2/27 (9x^2sin(3x)-2(-3xcos(3x)+sin(3x)))+C

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integral of 2x^2cos(3x)

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

\frac{2}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{27} d u

= 2 \cdot \int \frac{u ^{2} cos ( u )}{27} d u

Take the constant out:

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

= 2 \cdot \frac{1}{27} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{27} (u^{2} sin (u) - \int 2 u sin (u) d u)

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

= 2 \cdot \frac{1}{27} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 3 x

= 2 \cdot \frac{1}{27} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

Simplify

2 \cdot \frac{1}{27} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) : \frac{2}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

= \frac{2}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

Add a constant to the solution

= \frac{2}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + C

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What is the integral of 2x^2cos(3x) ?
The integral of 2x^2cos(3x) is 2/27 (9x^2sin(3x)-2(-3xcos(3x)+sin(3x)))+C