May 18, 2026 - Trigonometric Substitution Notes (Concise)

Fundamentals of Trigonometric Substitution

• Trigonometric substitution is used when integrals involve a square root of squared values, such as $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$.
• This technique converts algebraic expressions into trigonometric forms that can be solved using identities and standard integration methods learned in previous sessions.
• The process involves making a substitution for $x$, finding the derivative $dx$, simplifying the radical using Pythagorean identities, integrating with respect to $\theta$, and then back-substituting to the original variable $x$ using a reference triangle.

Case 1: Substitution for $\sqrt{a^2 - x^2}$

• Substitution: Let $x = a\sin(\theta)$.
• Differential: $dx = a\cos(\theta)\,d\theta$.
• Identity: Simplifies the expression using $a^2 - a^2\sin^2(\theta) = a^2(1 - \sin^2(\theta)) = a^2\cos^2(\theta)$.
• Reference Triangle: The opposite side is $x$ and the hypotenuse is $a$. The adjacent side is $\sqrt{a^2 - x^2}$.
• Example: For $\sqrt{9 - x^2}$, the substitution is $x = 3\sin(\theta)$. The domain of the radical requires $x$ to be between $-3$ and $3$, which corresponds to the range of the sine function.

Case 2: Substitution for $\sqrt{a^2 + x^2}$

• Substitution: Let $x = a\tan(\theta)$.
• Differential: $dx = a\sec^2(\theta)\,d\theta$.
• Identity: Simplifies the expression using $a^2 + a^2\tan^2(\theta) = a^2(1 + \tan^2(\theta)) = a^2\sec^2(\theta)$.
• Reference Triangle: The opposite side is $x$ and the adjacent side is $a$. The hypotenuse is $\sqrt{x^2 + a^2}$.
• Example: For $\sqrt{x^2 + 4}$, $a = 2$ and $x = 2\tan(\theta)$.

Case 3: Substitution for $\sqrt{x^2 - a^2}$

• Substitution: Let $x = a\sec(\theta)$.
• Differential: $dx = a\sec(\theta)\tan(\theta)\,d\theta$.
• Identity: Simplifies the expression using $a^2\sec^2(\theta) - a^2 = a^2(\sec^2(\theta) - 1) = a^2\tan^2(\theta)$.
• Reference Triangle: The hypotenuse is $x$ and the adjacent side is $a$. The opposite side is $\sqrt{x^2 - a^2}$.
• Example: For $\sqrt{x^2 - 2}$, where $a = \sqrt{2}$, the substitution is $x = \sqrt{2}\sec(\theta)$.

Integration and Simplification Techniques

• Power Reduction Formula: Often needed for even powers of sine or cosine; for example, $\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))$.
• U-Substitution: Frequently applied to resulting trigonometric integrals, such as letting $u = \cos(\theta)$ or $u = \sec(\theta)$.
• Definite Integrals: When bounds are present, they must be converted from $x$ values to $\theta$ values using the inverse trigonometric function (e.g., $\sec^{-1}(\frac{5}{3})$).
• Reduction and Elementary Forms: Integrals may result in forms like $\int \sec(\theta)\,d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$ or the integral for $\csc(\theta)$.

Questions & Discussion

• Student Question: Why did the term become $x^2 - 4$ in the example?
  • Response: The instructor corrected the value to $2$ because $a^2 = 2$, making $a = \sqrt{2}$. The triangle was updated accordingly.
• Student Question: Is the term supposed to be $5\sin^3(\theta)$ in the substitution for $x^3$?
  • Response: Yes, for a substitution where $x = 5\sin(\theta)$, $x^3$ becomes $5^3\sin^3(\theta) = 125\sin^3(\theta)$.
• Student Question: How to handle $\tan(\sec^{-1}(\frac{5}{3}))$?
  • Response: Use the reference triangle. If $\sec(\theta) = \frac{5}{3}$, then the hypotenuse is $5$ and the adjacent side is $3$. By the Pythagorean theorem ($3^2 + b^2 = 5^2$), the opposite side is $4$. Therefore, $\tan(\theta) = \frac{4}{3}$.
• General Note: Amber, Jordan, and Matthew were noted present. Quizzes will be returned tomorrow.