Calculus 2 2026-05-18 - Trigonometric Substitution Lecture

Introduction to Trigonometric Substitutions

• The Problem of the Square Root: In calculus, integrals involving square roots of quadratic expressions like $\sqrt{a^2 - x^2}$, $\sqrt{x^2 + a^2}$, or $\sqrt{x^2 - a^2}$ are difficult to solve using basic U-substitution or integration by parts.

• Geometric Motivation: If we consider a right triangle with a hypotenuse of $a$ and a leg of $x$, the Pythagorean theorem tells us the other side is $\sqrt{a^2 - x^2}$. This suggests we can express the variable $x$ in terms of trigonometric functions to simplify the integral.

• Domain and Range Considerations:

  • For an expression like $\sqrt{9 - x^2}$, the domain of $x$ is limited. If $x$ is too large or too small (e.g., $x = 5$ or $x = -5$), we get a negative value under the square root, resulting in a complex number. Thus, $x$ must be between $-3$ and $3$.

  • Because the range of $\sin(\theta)$ is between $-1$ and $1$, we can make the substitution $x = 3 \times \theta$ so that $-3 \times 1 \le x \le 3 \times 1$.

  • The Hidden Theorem: This substitution is valid due to the Intermediate Value Theorem, ensuring there is always a value of $\theta$ such that $x = a \times \theta$.

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

• The Transformation Identity: This case relies on the identity $\sin^2(\theta) + \text{cos}^2(\theta) = 1$, rearranged as $1 - \text{sin}^2(\theta) = \text{cos}^2(\theta)$.

• Substitution Rule: Set $x = a \text{ sin}(\theta)$.

• Differential Item: We must also change the differential: $dx = a \text{ cos}(\theta) d\theta$.

• Simplification Process (Example with $\sqrt{9 - x^2}$):

  1. Substitute $x = 3 \text{ sin}(\theta)$.

  2. $\sqrt{9 - (3 \text{ sin}(\theta))^2} = \sqrt{9 - 9 \text{ sin}^2(\theta)}$.

  3. Factor out the $9$: $\sqrt{9(1 - \text{sin}^2(\theta))} = 3 \sqrt{\text{cos}^2(\theta)}$.

  4. The square root is eliminated: $3 \text{ cos}(\theta)$.

• Integrating Powers of Cosine: If the substitution results in $\int \text{cos}^2(\theta) d\theta$, use the Power Reduction Formula (Double Angle Formula):

  • $\text{cos}^2(\theta) = \frac{1}{2}(1 + \text{cos}(2\theta))$

  • For sine: $\text{sin}^2(\theta) = \frac{1}{2}(1 - \text{cos}(2\theta))$

• Reversion to Original Variable: After integrating in terms of $\theta$, we must return to $x$.

  • If $x = 3 \text{ sin}(\theta)$, then $\theta = \text{sin}^{-1}(\frac{x}{3})$.

  • Use the reference triangle (SOH CAH TOA) to find functions like $\text{cos}(\theta)$ or $\text{tan}(\theta)$.

Example Walkthrough: $\int \frac{1}{x^2 \sqrt{4 - x^2}} dx$

• Identifying a: Here $a^2 = 4$, so $a = 2$.

• Setup:

  • $x = 2 \text{ sin}(\theta)$.

  • $dx = 2 \text{ cos}(\theta) d\theta$.

• Substitution: $\int \frac{2 \text{ cos}(\theta) d\theta}{(2 \text{ sin}(\theta))^2 \sqrt{4 - (2 \text{ sin}(\theta))^2}}$.

• Simplification:

  • Denominator: $4 \text{ sin}^2(\theta) \sqrt{4 - 4 \text{ sin}^2(\theta)} = 4 \text{ sin}^2(\theta) \times 2 \sqrt{1 - \text{sin}^2(\theta)} = 8 \text{ sin}^2(\theta) \text{ cos}(\theta)$.

  • The overall integral becomes $\int \frac{2 \text{ cos}(\theta)}{8 \text{ sin}^2(\theta) \text{ cos}(\theta)} d\theta$.

  • Cancellation: $\frac{1}{4} \int \frac{1}{\text{sin}^2(\theta)} d\theta = \frac{1}{4} \int \text{csc}^2(\theta) d\theta$.

• Integration: The elementary anti-derivative is $-\frac{1}{4} \text{ cot}(\theta) + C$.

• Back Substitution:

  • From our triangle: $\text{sin}(\theta) = \frac{x}{2}$ (opposite/hypotenuse).

  • Adjacent side = $\sqrt{2^2 - x^2} = \sqrt{4 - x^2}$.

  • $\text{cot}(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{4 - x^2}}{x}$.

  • Final Answer: $-\frac{\sqrt{4 - x^2}}{4x} + C$.

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

• Substitution Rule: Set $x = a \text{ tan}(\theta)$.

• Differential Item: $dx = a \text{ sec}^2(\theta) d\theta$.

• Transformation Identity: $1 + \text{tan}^2(\theta) = \text{sec}^2(\theta)$.

• Reference Triangle Geometry:

  • Opposite side: $x$.

  • Adjacent side: $a$.

  • Hypotenuse: $\sqrt{x^2 + a^2}$.

• Example Case: $\int \frac{x^3}{\sqrt{x^2 + 4}} dx$

  • $x = 2 \text{ tan}(\theta)$, $dx = 2 \text{ sec}^2(\theta) d\theta$.

  • $x^3 = 8 \text{ tan}^3(\theta)$.

  • The square root simplifies: $\sqrt{4 \text{ tan}^2(\theta) + 4} = 2 \text{ sec}(\theta)$.

  • Integral: $\int \frac{8 \text{ tan}^3(\theta) \cdot 2 \text{ sec}^2(\theta)}{2 \text{ sec}(\theta)} d\theta = 8 \int \text{ tan}^3(\theta) \text{ sec}(\theta) d\theta$.

  • Solving the Trig Integral: Use the identity $\text{tan}^2(\theta) = \text{sec}^2(\theta) - 1$.

  • $8 \int (\text{sec}^2(\theta) - 1) \text{ sec}(\theta) \text{ tan}(\theta) d\theta$.

  • Let $u = \text{sec}(\theta)$, $du = \text{sec}(\theta) \text{ tan}(\theta) d\theta$.

  • Integral becomes $8 \int (u^2 - 1) du = 8(\frac{u^3}{3} - u) + C$.

  • Revert to $x$ using the triangle: $\text{sec}(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{x^2 + 4}}{2}$.

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

• Substitution Rule: Set $x = a \text{ sec}(\theta)$.

• Differential Item: $dx = a \text{ sec}(\theta) \text{ tan}(\theta) d\theta$.

• Transformation Identity: $\text{sec}^2(\theta) - 1 = \text{tan}^2(\theta)$.

• Reference Triangle Geometry:

  • Hypotenuse: $x$.

  • Adjacent side: $a$.

  • Opposite side: $\sqrt{x^2 - a^2}$.

• Handling Non-Perfect Squares: If you have $\sqrt{x^2 - 2}$, then $a^2 = 2$, so $a = \sqrt{2}$.

• Example Integration: $\int \frac{1}{\sqrt{x^2 - 2}} dx$

  1. $x = \sqrt{2} \text{ sec}(\theta)$.

  2. $dx = \sqrt{2} \text{ sec}(\theta) \text{ tan}(\theta) d\theta$.

  3. Substitution: $\int \frac{\sqrt{2} \text{ sec}(\theta) \text{ tan}(\theta)}{\sqrt{2 \text{ sec}^2(\theta) - 2}} d\theta = \int \frac{\sqrt{2} \text{ sec}(\theta) \text{ tan}(\theta)}{\sqrt{2} \text{ tan}(\theta)} d\theta = \int \text{ sec}(\theta) d\theta$.

  4. Anti-derivative: $\text{ln}|\text{sec}(\theta) + \text{tan}(\theta)| + C$.

  5. Convert back to $x$: $\text{sec}(\theta) = \frac{x}{\sqrt{2}}$, $\text{tan}(\theta) = \frac{\sqrt{x^2 - 2}}{\sqrt{2}}$.

Complex Definite Integral Example

• The Problem: Solving an integral with bounds (e.g., $x = 3$ to $x = 5$) using trig substitution.

• Procedure:

  • When $\theta$ was substituted for $x$, we change the bounds accordingly using the inverse function.

  • Example logic: If $x = a \text{ sec}(\theta)$ and you evaluate at a specific point, you can avoid using the triangle at the end by evaluating in terms of $\theta$.

• Evaluating Results:

  • Calculation: $9 \times \frac{4}{3} \times \frac{5}{3} + \frac{1}{2} \text{ ln}(\frac{5}{3} + \frac{4}{3}) - 9 \text{ ln}(...)$.

  • The terms might involve constants like $\frac{9}{2} - 9 = -\frac{9}{2}$.

  • Algebraic simplification: $\text{ln}(1) = 0$, so many terms may drop out when plugging in lower bounds like $x = 0$ or $\theta = 0$.

Questions & Discussion

• Question (Audience): "Why did it become x like $x - 4$ over here?"

• Response (Instructor): "Oh, sorry… twos. Yeah, it is $2^2$ not $2$. This changes the reference triangle… the $2$ is now a $9$, which tells me that my $a$ should be $3$."

• Question (Audience): Concern about the duration of the calculation.

• Response (Instructor): "That problem took 20 minutes. This is why we don't do these very often. The integration is the easy part; the definite integral evaluation is usually the most cluttered algebra part."

• Administrative Note: The instructor will return quizzes tomorrow. The next session will focus on more examples involving these three triangles without being told which case to use.