What is the integral of 10x^2cos(9x) ?

The integral of 10x^2cos(9x) is 10/729 (81x^2sin(9x)-2(-9xcos(9x)+sin(9x)))+C

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integral of 10x^2cos(9x)

\int 10 x^{2} cos (9 x) d x

\frac{10}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))) + C

Solution steps

\int 10 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

= 10 \cdot \int x^{2} cos (9 x) d x

Apply u-substitution

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

= 10 \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

= 10 \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

= 10 \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))

= 10 \cdot \frac{1}{729} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 9 x

= 10 \cdot \frac{1}{729} ((9 x)^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x)))

Simplify

10 \cdot \frac{1}{729} ((9 x)^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))) : \frac{10}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x)))

= \frac{10}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x)))

Add a constant to the solution

= \frac{10}{729} (81 x^{2} sin (9 x) - 2 (- 9 x cos (9 x) + sin (9 x))) + C

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What is the integral of 10x^2cos(9x) ?
The integral of 10x^2cos(9x) is 10/729 (81x^2sin(9x)-2(-9xcos(9x)+sin(9x)))+C