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

The integral of x^2cos(kx) is 1/(k^3)(k^2x^2sin(kx)-2(-kxcos(kx)+sin(kx)))+C

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integral of x^2cos(kx)

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

\frac{1}{k ^{3}} (k^{2} x^{2} sin (k x) - 2 (- k x cos (k x) + sin (k x))) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{k ^{3}} d u

= \int \frac{u ^{2} cos ( u )}{k ^{3}} d u

Take the constant out:

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

= \frac{1}{k ^{3}} \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}{k ^{3}} (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}{k ^{3}} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = k x

= \frac{1}{k ^{3}} ((k x)^{2} sin (k x) - 2 (- k x cos (k x) + sin (k x)))

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \frac{1}{k ^{3}} (k^{2} x^{2} sin (k x) - 2 (- k x cos (k x) + sin (k x)))

Add a constant to the solution

= \frac{1}{k ^{3}} (k^{2} x^{2} sin (k x) - 2 (- k x cos (k x) + sin (k x))) + C

y^{^{'}} - 2 x y = x, y (0) = 0

\int 2^{2 x + 1} d x

f (x) = 2 - 3 x^{2}, at x = 1

\frac{d}{d x} ((x^{3})^{2})

\int \frac{1}{a x + b x ^{2}} d x

What is the integral of x^2cos(kx) ?
The integral of x^2cos(kx) is 1/(k^3)(k^2x^2sin(kx)-2(-kxcos(kx)+sin(kx)))+C