Integration Techniques for Powers of Sine and Cosine

Integration of Trigonometric Powers

General Strategy for Odd Exponents

• Key Idea: When at least one trigonometric function has an odd exponent, the strategy is to "break off" one copy of the odd-powered function to facilitate a u-substitution.
• Steps:
  1. Isolate one factor of the function with the odd exponent (e.g., ext{sin } x or ext{cos } x ).
  2. Rewrite the remaining even power using a Pythagorean identity (e.g., ext{sin}^2 x = 1 - ext{cos}^2 x or ext{cos}^2 x = 1 - ext{sin}^2 x ).
  3. Perform a u-substitution: Let u be the other trigonometric function (e.g., if you isolated ext{sin } x , let u = ext{cos } x ; if you isolated ext{cos } x , let u = ext{sin } x ). The isolated factor will become part of du .
  4. Integrate the resulting polynomial in u . \
  5. Substitute back to the original variable.
• Why it works: An odd exponent (e.g., n ) means we can write ext{func}^n x = ext{func}^{n-1} x imes ext{func } x . Since n-1 is even, ext{func}^{n-1} x can always be expressed in terms of the other function using the Pythagorean identity, leaving the single ext{func } x for du . This is why odd exponents are considered "nice" cases.

Case 1: One Odd Exponent (e.g., ext{sin}^n x ext{cos}^m x where n is odd and m is even or zero)

• Example 1: ext{integral of } ext{cos}^2 x ext{ with respect to } x (Initial discussion implies ext{sin}^0 x )
  • Strategy mentioned: Break off one ext{cos } x . This led to an example where ext{cos}^2 x was rewritten as (1 - ext{sin}^2 x) ext{cos } x .
  • Substitution for such a case: Let u = ext{sin } x . Then du = ext{cos } x dx .
  • The integral becomes ext{integral of } (1 - u^2) du .
  • Integration: u - rac{u^3}{3} + C .
  • Substitute back: ext{sin } x - rac{ ext{sin}^3 x}{3} + C .
• Example 2: ext{integral of } ext{sin}^5 x ext{cos}^2 x dx
  • Focus on the odd exponent: ext{sin}^5 x .
  • Break off one copy: ext{sin } x . The integral becomes ext{integral of } ext{sin}^4 x ext{cos}^2 x ext{sin } x dx .
  • Rewrite even power: ext{sin}^4 x = ( ext{sin}^2 x)^2 = (1 - ext{cos}^2 x)^2 .
  • The integral becomes ext{integral of } (1 - ext{cos}^2 x)^2 ext{cos}^2 x ext{sin } x dx .
  • Substitution: Let v = ext{cos } x . Then dv = - ext{sin } x dx , so - ext{dv} = ext{sin } x dx .
  • The integral becomes ext{integral of } (1 - v^2)^2 v^2 (-dv) = - ext{integral of } (1 - 2v^2 + v^4) v^2 dv .
  • Simplify: - ext{integral of } (v^2 - 2v^4 + v^6) dv .
  • Integrate using power rule: -rac{v^3}{3} + rac{2v^5}{5} - rac{v^7}{7} + C .
  • Substitute back v = ext{cos } x : -rac{1}{3} ext{cos}^3 x + rac{2}{5} ext{cos}^5 x - rac{1}{7} ext{cos}^7 x + C .

Case 2: Both Exponents are Odd (e.g., ext{sin}^n x ext{cos}^m x where both n and m are odd)

• Strategy: You can choose either of the odd-powered functions to apply the strategy from Case 1. Both choices will lead to a correct but potentially different-looking antiderivative.
• Example: ext{integral of } ext{sin}^3 x ext{cos}^3 x dx
  • Option A: Focus on ext{sin}^3 x
    • Break off ext{sin } x : ext{integral of } ext{sin}^2 x ext{cos}^3 x ext{sin } x dx .
    • Rewrite ext{sin}^2 x = 1 - ext{cos}^2 x : ext{integral of } (1 - ext{cos}^2 x) ext{cos}^3 x ext{sin } x dx .
    • Substitution: Let w = ext{cos } x . Then dw = - ext{sin } x dx , so - ext{dw} = ext{sin } x dx .
    • The integral becomes ext{integral of } (1 - w^2) w^3 (-dw) = ext{integral of } (w^5 - w^3) dw .
    • Integrate: rac{w^6}{6} - rac{w^4}{4} + C . (Note: A negative from the initial part was pulled out and distributed, switching the order of terms compared to the example).
    • Substitute back: rac{1}{6} ext{cos}^6 x - rac{1}{4} ext{cos}^4 x + C .
  • Option B: Focus on ext{cos}^3 x
    • Break off ext{cos } x : ext{integral of } ext{sin}^3 x ext{cos}^2 x ext{cos } x dx .
    • Rewrite ext{cos}^2 x = 1 - ext{sin}^2 x : ext{integral of } ext{sin}^3 x (1 - ext{sin}^2 x) ext{cos } x dx .
    • Substitution: Let z = ext{sin } x . Then dz = ext{cos } x dx .
    • The integral becomes ext{integral of } z^3 (1 - z^2) dz = ext{integral of } (z^3 - z^5) dz .
    • Integrate: rac{z^4}{4} - rac{z^6}{6} + C .
    • Substitute back: rac{1}{4} ext{sin}^4 x - rac{1}{6} ext{sin}^6 x + C .
• Note on Different Answers: Both Option A and Option B yield correct antiderivatives. They are equivalent up to the arbitrary constant C . This means their derivatives are identical, even if their specific functional forms appear different.
  • Graphical Interpretation: If two functions have the same derivative, their graphs have the same slope at every corresponding x -value. They are essentially vertical shifts of each other. The +C accounts for this vertical shift.

Case 3: Both Exponents are Even (e.g., ext{sin}^n x ext{cos}^m x where both n and m are even)

• Problematic Case: The previous substitution strategy is not viable because there's no single factor left over to form du .
• Solution: Use Power-Reducing (Double Angle) Identities repeatedly.
  • ext{cos}^2 heta = rac{1 + ext{cos}(2 heta)}{2}
  • ext{sin}^2 heta = rac{1 - ext{cos}(2 heta)}{2}
• Purpose: These identities reduce the power of the trigonometric function from squared to the first power, making it integrable.
• Example 1: ext{integral of } ext{cos}^2 x dx
  • Apply identity: ext{integral of } rac{1 + ext{cos}(2x)}{2} dx .
  • Split and integrate: ext{integral of } rac{1}{2} dx + ext{integral of } rac{1}{2} ext{cos}(2x) dx .
  • First term: rac{x}{2} .
  • Second term: Use substitution ext{gamma } ( ext{greek letter}) = 2x , so d ext{gamma} = 2 dx ext{ and } dx = rac{1}{2} d ext{gamma} .
    • rac{1}{2} ext{integral of } ext{cos}( ext{gamma}) rac{1}{2} d ext{gamma} = rac{1}{4} ext{integral of } ext{cos}( ext{gamma}) d ext{gamma} = rac{1}{4} ext{sin}( ext{gamma}) .
  • Combine and substitute back: rac{x}{2} + rac{1}{4} ext{sin}(2x) + C .
• Instructor's Opinion: For ext{sin}^2 x or ext{cos}^2 x , integration by parts can also be used and is sometimes preferred, but the double-angle identity method is equally valid and preferred by some.
• Example 2: ext{integral of } ext{cos}^4 x dx
  • Step 1: Rewrite as squared terms: ext{integral of } ( ext{cos}^2 x)^2 dx .
  • Step 2: Apply double angle identity: ext{integral of } ext{left( } rac{1 + ext{cos}(2x)}{2} ext{right)}^2 dx . (Note: here heta = x )
  • Step 3: Expand the square: ext{integral of } rac{1 + 2 ext{cos}(2x) + ext{cos}^2(2x)}{4} dx .
  • Step 4: Distribute and identify terms: rac{1}{4} ext{integral of } 1 dx + rac{1}{2} ext{integral of } ext{cos}(2x) dx + rac{1}{4} ext{integral of } ext{cos}^2(2x) dx . (Note: Here, ext{cos}^2(2x) is a new term that needs further reduction).
  • Step 5: Apply double angle identity again for the remaining squared term: For ext{cos}^2(2x) , use ext{cos}^2 heta = rac{1 + ext{cos}(2 heta)}{2} with heta = 2x . \
    • So, ext{cos}^2(2x) = rac{1 + ext{cos}(2 imes 2x)}{2} = rac{1 + ext{cos}(4x)}{2} .
  • Step 6: Integrate the new terms: The integral rac{1}{4} ext{integral of } ext{cos}^2(2x) dx becomes rac{1}{4} ext{integral of } rac{1 + ext{cos}(4x)}{2} dx = rac{1}{8} ext{integral of } (1 + ext{cos}(4x)) dx .
    • This integrates to rac{1}{8} ext{left( } x + rac{ ext{sin}(4x)}{4} ext{right)} .
  • Step 7: Combine all results: Add the antiderivatives of all terms and add a single +C .
• Conclusion for Even Powers: Integrating even powers of trigonometric functions often requires multiple applications of the power-reducing identities, leading to longer and more complex calculations. This is why even powers are considered "horrible" or "annoying."
• General approach for Complex Even Powers (e.g., ext{integral of } ext{sin}^2 x ext{cos}^2 x dx ):
  • You will need to use both double angle identities. A common approach is to combine the terms first: ext{sin}^2 x ext{cos}^2 x = ( ext{sin } x ext{cos } x)^2 .
  • Recall ext{sin}(2x) = 2 ext{sin } x ext{cos } x , so ext{sin } x ext{cos } x = rac{ ext{sin}(2x)}{2} .
  • Therefore, ( ext{sin } x ext{cos } x)^2 = ext{left( } rac{ ext{sin}(2x)}{2} ext{right)}^2 = rac{ ext{sin}^2(2x)}{4} .
  • Now apply the double angle identity for ext{sin}^2 heta with heta = 2x . \
    • rac{1}{4} ext{integral of } rac{1 - ext{cos}(2 imes 2x)}{2} dx = rac{1}{8} ext{integral of } (1 - ext{cos}(4x)) dx . \
    • This can then be integrated easily: rac{1}{8} ext{left( } x - rac{ ext{sin}(4x)}{4} ext{right)} + C .

Arbitrary Constant ( +C )

• Concept: When finding an antiderivative, there is an infinite family of functions that have the same derivative. They differ only by an arbitrary constant C .
• Uniqueness: Antiderivatives are not unique. For example, x^2 and x^2 + 1 both have a derivative of 2x . Graphically, they represent the same function shifted vertically.
• Solving for C : If additional information (e.g., an initial condition, a point the function must pass through) is provided, you can solve for a specific C , thus finding a unique antiderivative from the family of possibilities.