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: $u - \frac{u^3}{3} + C$.
  • Substitute back: $ext{sin } x - \frac{ ext{sin}^3 x}{3} + C$.
• 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: $-\frac{v^3}{3} + \frac{2v^5}{5} - \frac{v^7}{7} + C$.
  • Substitute back $v = ext{cos } x$: $-\frac{1}{3} ext{cos}^3 x + \frac{2}{5} ext{cos}^5 x - \frac{1}{7} ext{cos}^7 x + C$.

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: $\frac{w^6}{6} - \frac{w^4}{4} + C$. (Note: A negative from the initial part was pulled out and distributed, switching the order of terms compared to the example).
    • Substitute back: $\frac{1}{6} ext{cos}^6 x - \frac{1}{4} ext{cos}^4 x + C$.
  • 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: $\frac{z^4}{4} - \frac{z^6}{6} + C$.
    • Substitute back: $\frac{1}{4} ext{sin}^4 x - \frac{1}{6} ext{sin}^6 x + C$.
• 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.
  • $ext{cos}^2 heta = \frac{1 + ext{cos}(2 heta)}{2}$
  • $ext{sin}^2 heta = \frac{1 - ext{cos}(2 heta)}{2}$
• 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: $ext{integral of } \frac{1 + ext{cos}(2x)}{2} dx$.
  • Split and integrate: $ext{integral of } \frac{1}{2} dx + ext{integral of } \frac{1}{2} ext{cos}(2x) dx$.
  • First term: $\frac{x}{2}$.
  • Second term: Use substitution $ext{gamma } ( ext{greek letter}) = 2x$, so $d ext{gamma} = 2 dx ext{ and } dx = \frac{1}{2} d ext{gamma}$.
    • $\frac{1}{2} ext{integral of } ext{cos}( ext{gamma}) \frac{1}{2} d ext{gamma} = \frac{1}{4} ext{integral of } ext{cos}( ext{gamma}) d ext{gamma} = \frac{1}{4} ext{sin}( ext{gamma})$.
  • Combine and substitute back: $\frac{x}{2} + \frac{1}{4} ext{sin}(2x) + C$.
• 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: $ext{integral of } ext{left( } \frac{1 + ext{cos}(2x)}{2} ext{right)}^2 dx$. (Note: here $heta = x$)
  • Step 3: Expand the square: $ext{integral of } \frac{1 + 2 ext{cos}(2x) + ext{cos}^2(2x)}{4} dx$.
  • Step 4: Distribute and identify terms: $\frac{1}{4} ext{integral of } 1 dx + \frac{1}{2} ext{integral of } ext{cos}(2x) dx + \frac{1}{4} ext{integral of } ext{cos}^2(2x) dx$. (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 $ext{cos}^2 heta = \frac{1 + ext{cos}(2 heta)}{2}$ with $heta = 2x$. \
    • So, $ext{cos}^2(2x) = \frac{1 + ext{cos}(2 imes 2x)}{2} = \frac{1 + ext{cos}(4x)}{2}$.
  • Step 6: Integrate the new terms: The integral $\frac{1}{4} ext{integral of } ext{cos}^2(2x) dx$ becomes $\frac{1}{4} ext{integral of } \frac{1 + ext{cos}(4x)}{2} dx = \frac{1}{8} ext{integral of } (1 + ext{cos}(4x)) dx$.
    • This integrates to $\frac{1}{8} ext{left( } x + \frac{ ext{sin}(4x)}{4} ext{right)}$.
  • 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 $ext{sin } x ext{cos } x = \frac{ ext{sin}(2x)}{2}$.
  • Therefore, $( ext{sin } x ext{cos } x)^2 = ext{left( } \frac{ ext{sin}(2x)}{2} ext{right)}^2 = \frac{ ext{sin}^2(2x)}{4}$.
  • Now apply the double angle identity for $ext{sin}^2 heta$ with $heta = 2x$. \
    • $\frac{1}{4} ext{integral of } \frac{1 - ext{cos}(2 imes 2x)}{2} dx = \frac{1}{8} ext{integral of } (1 - ext{cos}(4x)) dx$. \
    • This can then be integrated easily: $\frac{1}{8} ext{left( } x - \frac{ ext{sin}(4x)}{4} ext{right)} + C$.

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.