What is the integral of 2y^2cos(9y) ?

The integral of 2y^2cos(9y) is 2/729 (81y^2sin(9y)-2(-9ycos(9y)+sin(9y)))+C

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integral of 2y^2cos(9y)

\int 2 y^{2} cos (9 y) d y

\frac{2}{729} (81 y^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y))) + C

Solution steps

\int 2 y^{2} cos (9 y) d y

Take the constant out:

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

= 2 \cdot \int y^{2} cos (9 y) d y

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{729} d u

= 2 \cdot \int \frac{u ^{2} cos ( u )}{729} d u

Take the constant out:

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

= 2 \cdot \frac{1}{729} \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}{729} (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}{729} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 9 y

= 2 \cdot \frac{1}{729} ((9 y)^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y)))

Simplify

2 \cdot \frac{1}{729} ((9 y)^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y))) : \frac{2}{729} (81 y^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y)))

= \frac{2}{729} (81 y^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y)))

Add a constant to the solution

= \frac{2}{729} (81 y^{2} sin (9 y) - 2 (- 9 y cos (9 y) + sin (9 y))) + C

y^{5} + 3 x^{2} y^{2} + 5 x^{4} = 49, (- 1, 2)

(- 5.3) (- 1.5)

x \to 5 lim (5 x + ∣ x - 5 ∣)

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- t}

y^{^{'}} = (t y^{2} - 1) y, y (1) = \frac{1}{2}

What is the integral of 2y^2cos(9y) ?
The integral of 2y^2cos(9y) is 2/729 (81y^2sin(9y)-2(-9ycos(9y)+sin(9y)))+C