Antiderivative of f(x) = x · cos 3x

This problem asks to find the antiderivative F(x) of the function f(x) = x · cos 3x satisfying the condition F(0) = 0.

Problem

$f (x) = x \cdot cos 3 x, F (0) = 0$

Solution

Step 1

Apply the integration by parts formula:

\int u d v = u v - \int v d u

Step 2

Choose u = x (then du = dx) and dv = cos 3x dx (then v = (1/3)·sin 3x):

u = x, d u = d x

d v = cos 3 x d x, v = \frac{1}{3} sin 3 x

Step 3

Substitute into the integration by parts formula:

\int x \cdot cos 3 x d x = \frac{x \cdot sin 3 x}{3} - \frac{1}{3} \int sin 3 x d x

Step 4

Evaluate the remaining integral and obtain the general expression for the antiderivative:

F (x) = \frac{x \cdot sin 3 x}{3} + \frac{cos 3 x}{9} + C

Step 5

Find the constant C from the condition F(0) = 0. Substitute x = 0: cos 0 = 1, therefore:

F (0) = 0 + \frac{1}{9} + C = 0

Step 6

From this, C = −1/9. The required antiderivative has the form:

Answer

$F (x) = \frac{x \cdot sin 3 x}{3} + \frac{cos 3 x}{9} - \frac{1}{9}$

Frequently asked questions

How do I find the antiderivative of f(x) = x · cos 3x?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = x (the polynomial is reduced when differentiated) and dv = cos 3x dx (easily integrated to (1/3)·sin 3x).

Why is v = (1/3)·sin 3x and not simply sin 3x?

Because integrating ∫ cos 3x dx gives (1/3)·sin 3x. Check: the derivative of (1/3)·sin 3x is (1/3)·3·cos 3x = cos 3x. The factor 1/3 appears due to the coefficient of x inside the cosine.

Why is C = −1/9 in this problem?

From the condition F(0) = 0. At x = 0: (0/3)·sin 0 + (1/9)·cos 0 + C = 0 + 1/9 + C = 0, hence C = −1/9.

When is integration by parts used?

When the integrand is a product of different types of functions: polynomial × exponential, polynomial × trigonometric, exponential × trigonometric, logarithm, or inverse trigonometric. The classic sign: the integral is not taken directly but simplifies after applying the formula ∫ u dv = uv − ∫ v du.