Find the indefinite integral \int \left( 3x(x - 2) + 8\cos(4x) \right) dx. — JAMB UTME

Find the indefinite integral $\int \left( 3x(x - 2) + 8\cos(4x) \right) dx$ .

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

Expanding

3x(x - 2)

yields

3x^2 - 6x

. Integrating

3x^2

gives

x^3

, integrating

-6x

yields

-3x^2

, and integrating

8\cos(4x)

yields

2\sin(4x)

. Adding the arbitrary constant

C

produces

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

Expand the algebraic term inside the integrand.

3x(x - 2) = 3x^2 - 6x

, expressing the integral as

\int (3x^2 - 6x + 8\cos(4x)) dx

Expanding the product converts the term into standard polynomial form suitable for integration.

Integrate each component term using basic integration rules.

\int 3x^2 dx = x^3

\int -6x dx = -3x^2

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and trigonometric rule

\int \cos(kx) dx = \frac{\sin(kx)}{k}

Combine terms and attach the constant of integration.

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

An indefinite integral requires an arbitrary constant

C

to represent the full set of antiderivatives.

Indefinite Integration of Polynomial and Trigonometric Functions

Question