What is the indefinite integral \int (4x^3 - 6\cos(3x)) \, dx? — JAMB UTME

What is the indefinite integral $\int (4x^3 - 6\cos(3x)) \, dx$ ?

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

Integrating term-by-term,

\int 4x^3 \, dx = x^4

and

\int -6\cos(3x) \, dx = -2\sin(3x)

. Summing these and including the arbitrary constant

C

produces

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

Integrate the polynomial term

4x^3

using the power rule

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

\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4

The power rule for integration adds

1

to the exponent and divides by the new exponent.

Integrate the trigonometric term

-6\cos(3x)

using

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx)

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

Integrating cosine gives positive sine, divided by the coefficient of

x

Combine the results and append the constant of integration

C

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

An indefinite integral represents a family of functions and requires the constant of integration

C

Indefinite Integration of Polynomial and Trigonometric Functions

Question