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

The integral of x^2cos(9x) is 1/729 (81x^2sin(9x)-2(-9xcos(9x)+sin(9x)))+C

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integral of x^2cos(9x)

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

\frac{1}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))) + C

Solution steps

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

Apply u-substitution

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

= \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

= \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

= \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))

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

Substitute back

u = 9 x

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

Simplify

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

= \frac{1}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x)))

Add a constant to the solution

= \frac{1}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))) + C

\frac{1}{5 s}

\frac{\partial}{\partial x} (x yz e^{x^{2} + y^{2} + z^{2}})

\int x 4 a^{2} - x^{2} d x

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2} + x y)

x \to \infty lim (- 1 x)

What is the integral of x^2cos(9x) ?
The integral of x^2cos(9x) is 1/729 (81x^2sin(9x)-2(-9xcos(9x)+sin(9x)))+C