What is the integral of x^2cos(6x) ?

The integral of x^2cos(6x) is 1/108 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C

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integral of x^2cos(6x)

\int x^{2} cos (6 x) d x

\frac{1}{108} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

Solution steps

\int x^{2} cos (6 x) d x

Apply u-substitution

= \int \frac{u ^{2} cos ( u )}{216} d u

Take the constant out:

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

= \frac{1}{216} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

= \frac{1}{216} (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}{216} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 6 x

= \frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x)))

Simplify

\frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x))) : \frac{1}{108} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

= \frac{1}{108} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

Add a constant to the solution

= \frac{1}{108} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

\int 5 - 4 x d x

\int_{0}^{1} \frac{1}{1 + x ^{2}} d x

\int e^{1 - 3 x} d x

y^{^{''}} + 2 y = e x + x^{3}

x \to 9 + lim (\frac{1}{x - 9})

What is the integral of x^2cos(6x) ?
The integral of x^2cos(6x) is 1/108 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C