May 19, 2026 - Study Notes on Trigonometric Substitution and Rational Function Integration

Fundamental Trigonometric Identities and the Unit Circle

• Core Pythagorean Identity: The fundamental identity is $\sin^2(\theta) + \cos^2(\theta) = 1$.
• Derived Pythagorean Identities:
  • Dividing by $\cos^2(\theta)$ yields: $\tan^2(\theta) + 1 = \sec^2(\theta)$.
  • Dividing by $\sin^2(\theta)$ yields: $1 + \cot^2(\theta) = \csc^2(\theta)$.
• Reciprocal Identities:
  • $\csc(\theta) = \frac{1}{\sin(\theta)}$
  • $\sec(\theta) = \frac{1}{\cos(\theta)}$
  • $\cot(\theta) = \frac{1}{\tan(\theta)}$
• Unit Circle Reference Values (First Quadrant):
  • At $\theta = 0$: $\cos(0) = 1$, $\sin(0) = 0$
  • At $\theta = \frac{\pi}{6}$: $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • At $\theta = \frac{\pi}{4}$: $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
  • At $\theta = \frac{\pi}{3}$: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
  • At $\theta = \frac{\pi}{2}$: $\cos\left(\frac{\pi}{2}\right) = 0$, $\sin\left(\frac{\pi}{2}\right) = 1$
• Common Anti-derivatives:
  • $\int \csc(\theta) \, d\theta = -\ln|\csc(\theta) + \cot(\theta)| + C$
  • The derivative of $\tan(\theta)$ is $\sec^2(\theta)$.

Techniques for Trigonometric Substitution

• Identifying Substitution Types:
  • Forms with Addition ($a^2 + x^2$): Use the tangent substitution ($x = a \tan(\theta)$).
  • Forms involving ($a^2 - x^2$): Use the sine substitution ($x = a \sin(\theta)$).
  • Forms involving ($x^2 - a^2$): Use the secant substitution ($x = a \sec(\theta)$).

Comprehensive Example: Integrating $\sqrt{1+4x^2}$

• Problem Setup: We are looking at the expression $\sqrt{1 + 4x^2}$. This fits the form $1 + u^2$, where $u = 2x$.
• Substitution: Let $2x = \tan(\theta)$. Then:
  • $x = \frac{1}{2} \tan(\theta)$
  • $dx = \frac{1}{2} \sec^2(\theta) \, d\theta$
• Transforming the Integral:
  • The expression $\sqrt{1 + (2x)^2}$ becomes $\sqrt{1 + \tan^2(\theta)} = \sqrt{\sec^2(\theta)} = \sec(\theta)$.
  • After performing the substitution and taking a specific case involving higher powers (like an raised power 4 in the transcript example), the coefficient becomes $\frac{1}{32}$.
• Power Reduction Formula for Secant:
  • $\int \sec^n(\theta) \, d\theta = \frac{1}{n-1} \sec^{n-2}(\theta) \tan(\theta) + \frac{n-2}{n-1} \int \sec^{n-2}(\theta) \, d\theta$
• Strategic Simplification: When facing a polynomial of secants (e.g., $\int (\sec^7(\theta) - 2\sec^5(\theta) + \sec^3(\theta)) \, d\theta$), it is best to apply the reduction formula to the highest power first. This often generates terms that can be combined with the lower-power integrals already present, reducing the overall workload.
• Example Combination of Coefficients: For a term involving $\frac{5}{6}$ and another involving $2$, we find the common denominator: $\frac{5}{6} + \frac{12}{6} = \frac{17}{6}$.
• Back-substitution using Reference Triangles:
  • If $\tan(\theta) = 2x$, draw a right triangle where the opposite side is $2x$ and the adjacent side is $1$.
  • The hypotenuse is $\sqrt{1 + 4x^2}$.
  • This allows for the conversion of all trigonometric terms ($\sec(\theta)$, etc.) back into terms of $x$.

Comparison of Integration Strategies: Trig Substitution vs. u-substitution

• Clever Observation: In some cases where trigonometric substitution seems necessary (like $\int x^3 \sqrt{x^2+1} \, d\theta$), a standard $u$-substitution can be more efficient.
• Substitution Strategy: Let $u = x^2 + 1$. Then $du = 2x \, dx$.
• Rewriting the Integral:
  • $x^3 = x^2 \times x$
  • Since $x^2 = u - 1$, the integral becomes $\int (u - 1) \sqrt{u} \frac{1}{2} \, du$.
  • Expand to $\frac{1}{2} \int (u^{3/2} - u^{1/2}) \, du$.
• Results comparison:
  • Power rule integration yields $\frac{1}{2} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) + C$.
  • This simplifies to $\frac{1}{5}(1+x^2)^{5/2} - \frac{1}{3}(1+x^2)^{3/2} + C$.
  • This method is significantly faster and less prone to error than managing multiple trigonometric power reductions.

Solving Integrals with Radical Forms via Trigonometric Simplification

• Example Problem: $\int \frac{\sqrt{x^2 - 16}}{x^4} \, dx$
  • Substitution Selection: Since it is $x^2 - a^2$, use $x = 4 \sec(\theta)$.
  • Calculations: $dx = 4 \sec(\theta) \tan(\theta) \, d\theta$.
  • Triangle: Hypotenuse is $x$, adjacent is $4$, opposite is $\sqrt{x^2 - 16}$.
  • Simplification Process: The expression simplifies into products and quotients of sine and cosine.
  • Resulting form: The integral may simplify to something like $\int \sin^2(u) \cos(u) \, du$, which is then solved via a simple substitution $u = \sin(\theta)$.
• Example Problem with Sine Substitution: $\int x^2 \sqrt{81 - x^2} \, dx$
  • Substitution: $x = 9 \sin(\theta)$, $dx = 9 \cos(\theta) \, d\theta$.
  • Simplification: Using the double-angle formula for $\cos^2(\theta)$: $\cos^2(\theta) = \frac{1}{2} (1 + \cos(2\theta))$.
  • Integration: $\int \frac{1}{2}(1 + \cos(2\theta)) \, d\theta = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C$.
  • Double Angle Identity: $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$. This allows one to use the reference triangle to revert to $x$.

Integration of Rational Functions via Polynomial Long Division

• Rule of Thumb: If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first.
• Case 1: $\frac{x^2 + 3x + 5}{x + 1}$
  • Dividing $x^2+3x+5$ by $x+1$ gives a quotient of $x+2$ and a remainder of $3$.
  • Rewrite as: $\int (x + 2 + \frac{3}{x+1}) \, dx$.
  • Integral: $\frac{x^2}{2} + 2x + 3\ln|x+1| + C$.
• Case 2: $\frac{x-3}{x+2}$
  • Divide to get $1 - \frac{5}{x+2}$.
  • Integral: $x - 5\ln|x+2| + C$.

Integration via Partial Fraction Decomposition

• Procedure:
  1. Factor the denominator completely.
  2. Set up the decomposition based on factors. For distinct linear factors: $\frac{P(x)}{Q(x)} = \frac{A}{ax+b} + \frac{B}{cx+d} + \dots$
• Example: $\int \frac{3x+2}{x^3 - x^2 - 2x} \, dx$
  • Factor Denominator: $x(x^2 - x - 2) = x(x-2)(x+1)$.
  • Set up Equation: $\frac{3x+2}{x(x-2)(x+1)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1}$.
  • Clear Denominators: $3x+2 = A(x-2)(x+1) + B(x)(x+1) + C(x)(x-2)$.
• Solving for Coefficients (Method of Plugging in Zeros):
  • Let $x=0$: $2 = A(-2)(1) \implies A = -1$.
  • Let $x=2$: $3(2)+2 = B(2)(3) \implies 8 = 6B \implies B = \frac{4}{3}$.
  • Let $x=-1$: $3(-1)+2 = C(-1)(-3) \implies -1 = 3C \implies C = -\frac{1}{3}$.
• Final Integration: $\int (-\frac{1}{x} + \frac{4/3}{x-2} - \frac{1/3}{x+1}) \, dx = -\ln|x| + \frac{4}{3}\ln|x-2| - \frac{1}{3}\ln|x+1| + C$.

Advanced Scenarios and Irreducible Quadratic Factors

• Irreducible Quadratics: If a factor is of the form $x^2 + a^2$ and cannot be factored into real linear factors, the numerator in the decomposition must be a linear term: $\frac{Ax+B}{x^2+a^2}$.
• Expected Anti-derivatives: These forms typically result in a combination of natural logarithms (via $u$-substitution) and arc-tangent functions ($\arctan(x)$).

Questions & Discussion

• Student Question: Will the test specify which method to use, or can we choose any method that works?
• Answer: Generally, it is "solve this and whatever you come up with." However, the instructor may occasionally specify a method like "Use Trig Substitution" or "Integration by Parts" to ensure mastery of specific techniques.
• Student Question: Regarding long division, does it always happen if the degree is the same or higher?
• Answer: Yes, if the degree of the numerator is higher in or equal to the denominator, division is the first step.