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CIE A-Level Maths Study Notes

2.5.2 Trigonometric Integrations

Trigonometric integrations play a pivotal role when integrating functions with squared trigonometric terms. This section sheds light on the techniques for integrating such functions, focusing on the use of trigonometric identities like sin2(x)sin²(x) and cos2(2x)cos²(2x).

Utilising Trigonometric Identities

The integration of trigonometric functions often necessitates the application of identities to simplify the expressions. Double-angle formulas are particularly useful for converting squared trigonometric functions into forms more amenable to integration.

Double-Angle Formulas

The relevant double-angle identities are:

sin2(x)=12(1cos(2x))sin²(x) = \frac{1}{2}(1 - cos(2x))cos2(x)=12(1+cos(2x))cos²(x) = \frac{1}{2}(1 + cos(2x))

These can be utilised to simplify integrals of sin2(x)sin²(x) and cos2(x)cos²(x).

Practical Example:

How do you calculate the area under the curve y=sin3(2x)cos3(2x)y = \sin^3(2x)\cos^3(2x)from x=0x = 0 to x=π2x = \frac{\pi}{2}?

Solution:

1. Identify the limits of integration:

  • The function intersects the x-axis at x=0x = 0 and x=π2x = \frac{\pi}{2}.

2. Implement the substitution u=sin(2x)u = sin(2x):

  • The differential is du=2cos(2x)dxdu = 2cos(2x)dx, implying dx=du2cos(2x)dx = \frac{du}{2cos(2x)}.

3. Simplify the expression and integrate:

  • The integral simplifies to sin3(2x)cos2(2x)du2cos(2x)\int sin³(2x)cos²(2x) \cdot \frac{du}{2cos(2x)}, which reduces to 12u3(1u2)du\frac{1}{2} \int u³(1 - u²) du.

4. Calculate the integral:

  • The result of the integration is u612+u48-\frac{u^6}{12} + \frac{u^4}{8}.

5. Apply the definite limits in terms of uu:

  • At x=0x = 0, u=0u = 0, and at x=π2x = \frac{\pi}{2}, u=1u = 1.
  • Evaluating the definite integral from u=0u = 0 to u=1u = 1, we get 00, since the positive and negative areas cancel each other out.

Therefore, the area under the curve y=sin3(2x)cos3(2x)y = sin³(2x)cos³(2x), denoted by AA, is confirmed to be 00.

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