Trigonometric Integration and De Moivre’s Theorem

Problem Statement

Integrating simple trigonometric functions like $\sin(x)$ and $\cos(x)$ is straightforward. However, when you encounter high powers of these functions—like $\sin^6(x)$ or $\cos^4(x)$—standard substitution fails. To solve these, you must rely on trigonometric identities (like the double angle formula) to reduce the power.

For very high powers, even standard trig identities become tedious. This is where the magic of complex numbers enters the picture. De Moivre’s Theorem can be used to elegantly prove trigonometric identities, converting high powers of $\sin$ or $\cos$ into a sum of multiple angles, which can then be easily integrated.

Consider the integral of $\cos^4(\theta)$. We will solve this using the connection between complex numbers and trigonometry. Let $z = \cos\theta + i\sin\theta$.

(a) Show that $z + z^{-1} = 2\cos\theta$ and $z^n + z^{-n} = 2\cos(n\theta)$.

(b) By expanding $(z + z^{-1})^4$ using the binomial theorem, express $\cos^4(\theta)$ in the form $A \cos(4\theta) + B \cos(2\theta) + C$.

(c) Hence, evaluate the indefinite integral $\int \cos^4(\theta) \, d\theta$.

Hints

• Part (a): Use De Moivre's Theorem: $z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$. Remember that $\cos(-x) = \cos(x)$ and $\sin(-x) = -\sin(x)$.
• Part (b): The binomial expansion of $(A + B)^4$ is $A^4 + 4A^3B + 6A^2B^2 + 4AB^3 + B^4$. Substitute $z$ and $z^{-1}$, then group the terms into pairs of $(z^n + z^{-n})$. Use the result from part (a) to convert these back into cosines.
• Part (c): Substitute the expanded expression from part (b) into the integral. You are now integrating a linear combination of $\cos(4\theta)$ and $\cos(2\theta)$, which is a standard calculus procedure.

Solutions

Part (a): Complex Number Properties

1. Let $z = \cos\theta + i\sin\theta$.
2. By De Moivre's Theorem, $z^n = \cos(n\theta) + i\sin(n\theta)$.
3. Similarly, $z^{-n} = \cos(-n\theta) + i\sin(-n\theta)$.
4. Since cosine is an even function and sine is an odd function: $z^{-n} = \cos(n\theta) - i\sin(n\theta)$
5. Adding $z^n$ and $z^{-n}$ together: $z^n + z^{-n} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(n\theta) - i\sin(n\theta))$ $z^n + z^{-n} = 2\cos(n\theta)$
6. If $n=1$, we get the specific case: $z + z^{-1} = 2\cos\theta$

Part (b): Expanding with the Binomial Theorem

1. From (a), we know $(2\cos\theta)^4 = (z + z^{-1})^4$.
2. Expand using the binomial theorem (coefficients 1, 4, 6, 4, 1): $(z + z^{-1})^4 = z^4 + 4(z^3)(z^{-1}) + 6(z^2)(z^{-2}) + 4(z)(z^{-3}) + (z^{-1})^4$ $(z + z^{-1})^4 = z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4}$
3. Group the terms with matching powers: $(2\cos\theta)^4 = (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6$
4. Use the identity $z^n + z^{-n} = 2\cos(n\theta)$ to convert back to trigonometry: $16\cos^4\theta = 2\cos(4\theta) + 4(2\cos(2\theta)) + 6$ $16\cos^4\theta = 2\cos(4\theta) + 8\cos(2\theta) + 6$
5. Divide by 16 to isolate $\cos^4\theta$: $\cos^4\theta = \frac{1}{8}\cos(4\theta) + \frac{1}{2}\cos(2\theta) + \frac{3}{8}$ (This is in the required form $A \cos(4\theta) + B \cos(2\theta) + C$).

Part (c): Evaluating the Integral

1. Substitute the expanded expression into the integral: $\int \cos^4(\theta) \, d\theta = \int \left( \frac{1}{8}\cos(4\theta) + \frac{1}{2}\cos(2\theta) + \frac{3}{8} \right) \, d\theta$
2. Integrate term by term, remembering that $\int \cos(kx) dx = \frac{1}{k}\sin(kx)$:
   • $\int \frac{1}{8}\cos(4\theta) \, d\theta = \frac{1}{8} \times \frac{1}{4}\sin(4\theta) = \frac{1}{32}\sin(4\theta)$
   • $\int \frac{1}{2}\cos(2\theta) \, d\theta = \frac{1}{2} \times \frac{1}{2}\sin(2\theta) = \frac{1}{4}\sin(2\theta)$
   • $\int \frac{3}{8} \, d\theta = \frac{3}{8}\theta$
3. Combine and add the constant of integration $C$: $\int \cos^4(\theta) \, d\theta = \frac{1}{32}\sin(4\theta) + \frac{1}{4}\sin(2\theta) + \frac{3}{8}\theta + C$

Takeaways

• Cross-Topic Application: This problem highlights the beauty of Extension 2 Mathematics. You use Complex Numbers (De Moivre's) and Algebra (Binomial Theorem) to solve a Calculus (Integration) problem.
• Power Reduction: The overarching goal when integrating even powers of sine or cosine is always "power reduction." You must convert $\cos^n(\theta)$ into expressions involving $\cos(k\theta)$ with no powers.
• The $z + z^{-1}$ Trick: Memorise the identities $z + z^{-1} = 2\cos\theta$ and $z - z^{-1} = 2i\sin\theta$. They are your keys to unlocking these high-power trigonometric integrals.

Further Readings

• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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