What is the result of evaluating the indefinite integral \int \left( (2x - 1)^2 - 6\cos(2x… — JAMB UTME

What is the result of evaluating the indefinite integral $\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx$ ?

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Expanding the squared term

(2x - 1)^2

yields

4x^2 - 4x + 1

. Integrating term by term using the power rule gives

\frac{4}{3}x^3 - 2x^2 + x

. The integral of

-6\cos(2x)

-\frac{6}{2}\sin(2x) = -3\sin(2x)

. Adding the arbitrary constant

C

results in

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Expand the squared binomial term inside the integral

(2x - 1)^2 = 4x^2 - 4x + 1

Expanding the expression allows for term-by-term integration using standard integration rules.

Integrate each term of the expanded polynomial

\int (4x^2 - 4x + 1) \, dx = \frac{4}{3}x^3 - 2x^2 + x

Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.

Integrate the trigonometric term

\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)

The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).

Combine all integrated terms and append the constant of integration

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Indefinite integration requires an arbitrary constant of integration C.

Indefinite Integration of Polynomial and Trigonometric Functions

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