How do I find the antiderivative of f(x) = x · sin 2x?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = x (the polynomial lowers its degree when differentiated) and dv = sin 2x dx (easily integrated to −(1/2)·cos 2x).

Why is v = −(1/2)·cos 2x and not simply −cos 2x?

Because integrating ∫ sin 2x dx gives −(1/2)·cos 2x. Check: the derivative of −(1/2)·cos 2x is (1/2)·2·sin 2x = sin 2x. The factor 1/2 appears due to the coefficient of x inside the sine.

Why is C = 0 in this problem?

From the condition F(0) = 0. At x = 0: sin 0 = 0, therefore F(0) = 0 − 0 + C = 0, hence C = 0.

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.

Antiderivative of f(x) = x · sin 2x

This problem asks to find the antiderivative F(x) of the function f(x) = x · sin 2x satisfying the condition F(0) = 0.

Problem

$f (x) = x \cdot sin 2 x, F (0) = 0$

Solution

Step 1

The integrand x · sin 2x is a product of a polynomial and a trigonometric function. Such integrals are evaluated by integration by parts.

Step 2

Split the integrand x · sin 2x dx into two parts: u = x and dv = sin 2x dx. Then du = dx, and v is found by integrating dv: v = −(1/2)·cos 2x.

Why is the integration by parts method chosen?

The integrand is a product of a polynomial (x) and a trigonometric function (sin 2x). Neither substitution nor tabular formulas work here. The method of integration by parts is specifically designed for such integrals. The idea: by differentiating the polynomial, we lower its degree, and the integral gradually simplifies to a tabular one.

Why is u = x and not u = sin 2x?

In integration by parts, u is chosen to be the function that simplifies upon differentiation. If u = x, then du = dx — the polynomial lowers its degree. If u = sin 2x, then du = 2cos 2x dx — the trigonometric function does not simplify but remains equally complex. According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), the polynomial comes before the trigonometric function, so u = x.

Why is dv = sin 2x dx and not simply sin 2x?

The integration by parts formula: ∫ u dv = uv − ∫ v du. Here dv is the differential of the function v, and by definition dv = v′(x) dx. Without dx this is not a differential but a function. The product u · dv = x · sin 2x dx gives the original integrand.

u = x, d u = d x

d v = sin 2 x d x, v = - \frac{1}{2} cos 2 x

Step 3

Substitute into the integration by parts formula:

\int x \cdot sin 2 x d x = - \frac{x \cdot cos 2 x}{2} + \frac{1}{2} \int cos 2 x d x

Step 4

Evaluate the remaining integral ∫ cos 2x dx = (1/2)·sin 2x and obtain the general expression:

F (x) = - \frac{x \cdot cos 2 x}{2} + \frac{sin 2 x}{4} + C

Step 5

Find the constant C from the condition F(0) = 0. sin 0 = 0, therefore:

F (0) = 0 + \frac{sin 0}{4} + C = 0

Step 6

From this, C = 0. The required antiderivative has the form:

Answer

$F (x) = - \frac{x \cdot cos 2 x}{2} + \frac{sin 2 x}{4}$