What is the integral of 27x^2cos(3x) ?

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

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integral of 27x^2cos(3x)

\int 27 x^{2} cos (3 x) d x

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

Solution steps

\int 27 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

= 27 \cdot \int x^{2} cos (3 x) d x

Apply u-substitution

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

= 27 \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

= 27 \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

= 27 \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))

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

Substitute back

u = 3 x

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

Simplify

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

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

Add a constant to the solution

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

(5, - 5), (1, 5)

\int - \frac{1000}{x ^{2}} d x

\frac{d y}{d x} = \frac{1}{2} y

y = \frac{4 x ^{2} + 4 x + 2}{x}

\int_{0}^{3} \frac{1}{3 - x} d x

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