Math  /  Calculus

Question2. x2cos(3x)dx\int x^{2} \cos (3 x) d x

Studdy Solution

STEP 1

1. The integral x2cos(3x)dx\int x^{2} \cos(3x) \, dx is a product of a polynomial and a trigonometric function.
2. We will use integration by parts, which is suitable for integrals of the form udv\int u \, dv.

STEP 2

1. Identify uu and dvdv for integration by parts.
2. Differentiate uu and integrate dvdv.
3. Apply the integration by parts formula.
4. Simplify and solve the resulting integral.

STEP 3

Identify the parts for integration by parts. Let: u=x2 u = x^2 dv=cos(3x)dx dv = \cos(3x) \, dx

STEP 4

Differentiate uu and integrate dvdv: du=2xdx du = 2x \, dx v=cos(3x)dx=13sin(3x) v = \int \cos(3x) \, dx = \frac{1}{3} \sin(3x)

STEP 5

Apply the integration by parts formula udv=uvvdu\int u \, dv = uv - \int v \, du: x2cos(3x)dx=x213sin(3x)13sin(3x)2xdx\int x^2 \cos(3x) \, dx = x^2 \cdot \frac{1}{3} \sin(3x) - \int \frac{1}{3} \sin(3x) \cdot 2x \, dx =13x2sin(3x)23xsin(3x)dx= \frac{1}{3} x^2 \sin(3x) - \frac{2}{3} \int x \sin(3x) \, dx

STEP 6

For the remaining integral xsin(3x)dx\int x \sin(3x) \, dx, apply integration by parts again. Let: u=x u = x dv=sin(3x)dx dv = \sin(3x) \, dx
Differentiate uu and integrate dvdv: du=dx du = dx v=sin(3x)dx=13cos(3x) v = \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x)
Apply the integration by parts formula: xsin(3x)dx=x(13cos(3x))(13cos(3x))dx\int x \sin(3x) \, dx = x \cdot \left(-\frac{1}{3} \cos(3x)\right) - \int \left(-\frac{1}{3} \cos(3x)\right) \, dx =13xcos(3x)+13cos(3x)dx= -\frac{1}{3} x \cos(3x) + \frac{1}{3} \int \cos(3x) \, dx =13xcos(3x)+19sin(3x)= -\frac{1}{3} x \cos(3x) + \frac{1}{9} \sin(3x)

STEP 7

Substitute back to solve the original integral: x2cos(3x)dx=13x2sin(3x)23(13xcos(3x)+19sin(3x))\int x^2 \cos(3x) \, dx = \frac{1}{3} x^2 \sin(3x) - \frac{2}{3} \left(-\frac{1}{3} x \cos(3x) + \frac{1}{9} \sin(3x)\right) =13x2sin(3x)+29xcos(3x)227sin(3x)+C= \frac{1}{3} x^2 \sin(3x) + \frac{2}{9} x \cos(3x) - \frac{2}{27} \sin(3x) + C
where C C is the constant of integration.
The solution to the integral is:
13x2sin(3x)+29xcos(3x)227sin(3x)+C\boxed{\frac{1}{3} x^2 \sin(3x) + \frac{2}{9} x \cos(3x) - \frac{2}{27} \sin(3x) + C}

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