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Difficulty: Very hardIndefinite Integration of Polynomial and Trigonometric Functions

What is the indefinite integral ∫((2x−3)2+12sin⁡(4x)−6cos⁡(3x))dx\int \left( (2x - 3)^2 + 12\sin(4x) - 6\cos(3x) \right) dx?

  1. 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + CAnswer
  2. B
    43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x)
  3. C
    43x3−6x2+9x+3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x + 3\cos(4x) - 2\sin(3x) + C
  4. D
    43x3−6x2+9x−48cos⁡(4x)+18sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 48\cos(4x) + 18\sin(3x) + C

Answer

43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Expanding (2x−3)2(2x-3)^2 yields 4x2−12x+94x^2 - 12x + 9. Integrating term by term: ∫4x2dx=43x3\int 4x^2 dx = \frac{4}{3}x^3, ∫−12xdx=−6x2\int -12x dx = -6x^2, ∫9dx=9x\int 9 dx = 9x, ∫12sin⁡(4x)dx=−3cos⁡(4x)\int 12\sin(4x) dx = -3\cos(4x), and ∫−6cos⁡(3x)dx=−2sin⁡(3x)\int -6\cos(3x) dx = -2\sin(3x). Summing these results and adding the constant of integration CC gives 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C.

Step-by-Step Solution

1
Expand the squared binomial inside the integrand.
(2x−3)2=4x2−12x+9(2x - 3)^2 = 4x^2 - 12x + 9
Expanding algebraic terms simplifies the integrand into basic power terms ready for integration.
2
Integrate the polynomial terms using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
\int (4x^2 - 12x + 9) dx = \frac{4}{3}x^3 - 6x^2 + 9x
Applying the power rule to each term individually gives 4x33−12x22+9x\frac{4x^3}{3} - \frac{12x^2}{2} + 9x.
3
Integrate the trigonometric terms using standard integration rules ∫sin⁡(kx)dx=−1kcos⁡(kx)\int \sin(kx) dx = -\frac{1}{k}\cos(kx) and ∫cos⁡(kx)dx=1ksin⁡(kx)\int \cos(kx) dx = \frac{1}{k}\sin(kx).
\int 12\sin(4x) dx = -\frac{12}{4}\cos(4x) = -3\cos(4x) and \int -6\cos(3x) dx = -\frac{6}{3}\sin(3x) = -2\sin(3x)
Integrating sine gives negative cosine divided by the inner linear coefficient, and integrating cosine gives sine divided by the inner linear coefficient.
4
Combine all integrated terms and append the constant of integration CC.
\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Indefinite integrals require the addition of an arbitrary constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
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Topics

Indefinite Integration of Polynomial and Trigonometric Functions

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