What is the integral of x^2cos(8x) ?

The integral of x^2cos(8x) is 1/256 (32x^2sin(8x)-sin(8x)+8xcos(8x))+C

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integral of x^2cos(8x)

\int x^{2} cos (8 x) d x

\frac{1}{256} (32 x^{2} sin (8 x) - sin (8 x) + 8 x cos (8 x)) + C

Solution steps

\int x^{2} cos (8 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{512} d u

= \int \frac{u ^{2} cos ( u )}{512} d u

Take the constant out:

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

= \frac{1}{512} \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}{512} (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}{512} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 8 x

= \frac{1}{512} ((8 x)^{2} sin (8 x) - 2 (- 8 x cos (8 x) + sin (8 x)))

Simplify

\frac{1}{512} ((8 x)^{2} sin (8 x) - 2 (- 8 x cos (8 x) + sin (8 x))) : \frac{1}{256} (32 x^{2} sin (8 x) - sin (8 x) + 8 x cos (8 x))

= \frac{1}{256} (32 x^{2} sin (8 x) - sin (8 x) + 8 x cos (8 x))

Add a constant to the solution

= \frac{1}{256} (32 x^{2} sin (8 x) - sin (8 x) + 8 x cos (8 x)) + C

f (x) = \frac{56}{x ^{9}}

x \to 0 lim (x^{2} - 3)

x \to 2 lim (\frac{x + 5}{x - 2})

\int 2 sin (x) - 1 d x

\frac{d}{d x} (sin^{5} (ln (e^{3 x - 1})))

What is the integral of x^2cos(8x) ?
The integral of x^2cos(8x) is 1/256 (32x^2sin(8x)-sin(8x)+8xcos(8x))+C