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CIE A-Level Maths Study Notes

2.5.4 Integrating Special Functions

Integration is allowing us to find areas, volumes, and solve numerous real-world problems. This section focuses on integrating special functions, specifically x2+axx^2 + ax and tan(x)\tan(x), which are prevalent in advanced mathematics. Understanding these integrations is crucial for solving complex calculus problems.

Integration Techniques

1. Integration of x2+axx^2 + ax

The function x2+axx^2 + ax represents a quadratic-linear form and is integrated by addressing each term individually.

  • For the term x2x^2, the integral is x33\frac{x^3}{3}.
  • For the term axax, the integral is ax22\frac{ax^2}{2}. Combining these, we obtain: (x2+ax)dx=x33+ax22+C\int (x^2 + ax) \, dx = \frac{x^3}{3} + \frac{ax^2}{2} + C where C represents the constant of integration.

2. Integration of tan(x)\tan(x)

Integrating the trigonometric function tan(x)\tan(x) can be intricate. Utilising the identity tan(x)=sin(x)cos(x)\tan(x) = \frac{\sin(x)}{\cos(x)}, we approach the integral involving a logarithmic function: tan(x)dx=lncos(x)+C.\int \tan(x) \, dx = -\ln|\cos(x)| + C. This result is derived through substitution methods and leveraging the properties of logarithmic functions.

Advanced Example Integrations

Example 1: Integrating x2+3xx^2 + 3x

  • Identify the Function: The function is x2+3xx^2 + 3x.
  • Apply Integration Rules:
    • Integrate x2x^2: Using xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}, we get x2dx=x33\int x^2 \, dx = \frac{x^3}{3}.
    • Integrate 3x3x: Applying the power rule,3xdx=3x22\int 3x \, dx = \frac{3x^2}{2}.
  • Combine Results: Thus, (x2+3x)dx=x33+3x22+C\int (x^2 + 3x) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + C

Example 2: Integrating tan(x)\tan(x)

  • Identify the Function: The function is tan(x)\tan(x).
  • Apply Integration Rules: The integral of tan(x)\tan(x) is known: tan(x)dx=lncos(x)+C\int \tan(x) \, dx = -\ln|\cos(x)| + C

Example 3: Integrating x2+4xx^2 + 4x

  • Identify the Function: The function is x2+4xx^2 + 4x.
  • Apply Integration Rules:
    • Integrate x2x^2: x2dx=x33\int x^2 \, dx = \frac{x^3}{3}.
    • Integrate 4x4x: 4xdx=2x2\int 4x \, dx = 2x^2.
  • Combine Results: Hence, (x2+4x)dx=x33+2x2+C\int (x^2 + 4x) \, dx = \frac{x^3}{3} + 2x^2 + C

Example 4: Integrating tan(2x)\tan(2x)

  • Identify the Function: The function is tan(2x)tan(2x).
  • Apply Integration Rules with a Substitution:
    • Let u=2xu = 2x, then du=2dxdu = 2dx or dx=du2dx = \frac{du}{2}.
    • Substitute into the integral: tan(2x)dx=tan(u)du2\int \tan(2x) \, dx = \int \tan(u) \cdot \frac{du}{2}
    • Now integrate: tan(u)du=lncos(u)\int \tan(u) \, du = -\ln|\cos(u)|.
  • Back-Substitute and Combine Results:
    • Replace uu with 2x:lncos(2x)2x: -\ln|\cos(2x)|.
    • Include the factor of 12\frac{1}{2} from the substitution: 12lncos(2x)+C-\frac{1}{2} \ln|\cos(2x)| + C.

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