What is the integral of 729x^2cos(9x) ?

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

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integral of 729x^2cos(9x)

\int 729 x^{2} cos (9 x) d x

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

Solution steps

\int 729 x^{2} cos (9 x) d x

Take the constant out:

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

= 729 \cdot \int x^{2} cos (9 x) d x

Apply u-substitution

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

= 729 \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

= 729 \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

= 729 \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))

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

Substitute back

u = 9 x

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

Simplify

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

= 81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))

Add a constant to the solution

= 81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x)) + C

\frac{d}{d x} (\frac{2 x}{x ^{2} + y ^{4}})

\frac{\partial}{\partial x} (x^{2} - x y^{3} + 1)

\frac{\partial}{\partial x} (x^{2} + 8 y^{2} + 9 z^{2})

y^{^{''}} - 2 y^{^{'}} - 3 y = - 7 t e^{- t}

ln (cos (x) + sin (x))

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