Calculus 2 2026-05-19 - Trigonometric Substitution and Integration Techniques

Pythagorean Identities:
- The primary identity is $\sin^2(\theta) + \cos^2(\theta) = 1$ .
- By dividing the primary identity by $\cos^2(\theta)$ , the second identity is derived: $\tan^2(\theta) + 1 = \sec^2(\theta)$ .
- By dividing the primary identity by $\sin^2(\theta)$ , the third identity is derived: $1 + \cot^2(\theta) = \csc^2(\theta)$ .
- These identities allow solving for specific squared terms, such as $\sin^2(\theta) = 1 - \cos^2(\theta)$ .
Reciprocal Identities:
- $\csc(\theta) = \frac{1}{\sin(\theta)}$
- $\sec(\theta) = \frac{1}{\cos(\theta)}$
- $\cot(\theta) = \frac{1}{\tan(\theta)}$
The Unit Circle (First Quadrant):
- At angle $0$ : $\cos(0) = 1$ , $\sin(0) = 0$ .
- At angle $\frac{\pi}{6}$ : $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ , $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ .
- At angle $\frac{\pi}{4}$ : both $\cos\left(\frac{\pi}{4}\right)$ and $\sin\left(\frac{\pi}{4}\right)$ are $\frac{\sqrt{2}}{2}$ .
- At angle $\frac{\pi}{3}$ : $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ , $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ .
- At angle $\frac{\pi}{2}$ : $\cos\left(\frac{\pi}{2}\right) = 0$ , $\sin\left(\frac{\pi}{2}\right) = 1$ .

Anti-derivatives of Logarithmic Trig Functions:
- $\int \sec(\theta) d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$
- $\int \csc(\theta) d\theta = -\ln|\csc(\theta) + \cot(\theta)| + C$
Trigonometric Substitution Selection Rules:
- If the expression contains $\sqrt{a^2 + x^2}$ , use $x = a\tan(\theta)$ .
- If the expression contains $\sqrt{a^2 - x^2}$ , use $x = a\sin(\theta)$ .
- If the expression contains $\sqrt{x^2 - a^2}$ , use $x = a\sec(\theta)$ .
Trigonometric Substitution Example Case Study:
- Consider the integral involving $\sqrt{1 + 4x^2}$ .
- Rewrite as $\sqrt{1 + (2x)^2}$ , resulting in the substitution $2x = \tan(\theta)$ .
- Corresponding differential: $dx = \frac{1}{2}\sec^2(\theta)d\theta$ .
- Substitution results in $\sqrt{\tan^2(\theta) + 1} = \sqrt{\sec^2(\theta)} = \sec(\theta)$ .
- During the process of solving, constants are accumulated (e.g., factors of $16 \times 2 = 32$ ) and powers are applied (e.g., raising terms to the power of $4$ ).

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 Approach to Complex Integrals:
- When faced with a "polynomial of secants" (e.g., terms of $\sec^7(\theta), \sec^5(\theta), \sec^3(\theta)$ ), address the largest power first.
- Reducing the largest power will often generate recursive terms that can be combined with existing lower-power integrals, reducing the total work required.
- Example: Combining the resulting $\sec^5(\theta)$ integral from reducing $\sec^7(\theta)$ with the original $\sec^5(\theta)$ term using coefficients like $\frac{17}{6}$ .
Final Substitution Back to x:
- Always return the expression to the original variable $x$ using a reference triangle.
- If $2x = \tan(\theta)$ , then the opposite side is $2x$ , the adjacent side is $1$ , and the hypotenuse is $\sqrt{1 + 4x^2}$ .
- Thus, $\sec(\theta) = \sqrt{1 + 4x^2}$ .

Philosophy of "Winning" in Math:
- Math is compared to a game; the goal is to find the anti-derivative efficiently.
- Trig substitution is an "overpowered tool," but if a standard u-substitution works, it is much faster.
Comparison Example (U-sub vs. Trig Sub):
- Problem: $\int x^3 \sqrt{x^2 + 1} dx$ .
- Method A (Trig Sub): Substitute $x = \tan(\theta)$ . This leads to complex powers of secants and tangents.
- Method B (U-sub): Let $u = x^2 + 1$ , meaning $du = 2x dx$ .
- Rewrite $x^3 dx$ as $x^2 (x dx) = (u - 1) \frac{du}{2}$ .
- The integral becomes $\frac{1}{2} \int (u - 1) \sqrt{u} du = \frac{1}{2} \int (u^{3/2} - u^{1/2}) du$ .
- Solution: $\frac{1}{2} \left( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} \right) + C = \frac{1}{5}(1+x^2)^{5/2} - \frac{1}{3}(1+x^2)^{3/2} + C$ .
- Lesson: Check if the power outside the radical is one less than the power inside; if so, u-substitution is preferable.

Case 1: Degree of Numerator $\geq$ Degree of Denominator:
- In these cases, you MUST perform polynomial long division first.
- The integral is rewritten as the $\text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}}$ .
- Example: $\frac{x^2 + 3x + 5}{x + 1} = (x + 2) + \frac{3}{x + 1}$ .
- Integrating this yields $\frac{1}{2}x^2 + 2x + 3\ln|x + 1| + C$ .
Case 2: Partial Fraction Decomposition (PFD):
- Used when the denominator is factorable and the degree of the numerator is less than the denominator.
- Example Problem: $\int \frac{3x + 2}{x^3 - x^2 - 2x} dx$ .
- Step 1: Factor the denominator: $x(x^2 - x - 2) = x(x - 2)(x + 1)$ .
- Step 2: Set up decomposition: $\frac{3x + 2}{x(x - 2)(x + 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}$ .
- Step 3: Solve for coefficients $A, B, C$ using the "Heaviside" or plugging-in method:
  - Multiply by the common denominator: $3x + 2 = A(x - 2)(x + 1) + B(x)(x + 1) + C(x)(x - 2)$ .
  - To find $A$ , set $x = 0$ : $2 = A(-2)(1) \implies A = -1$ .
  - To find $B$ , set $x = 2$ : $3(2) + 2 = B(2)(2 + 1) \implies 8 = 6B \implies B = \frac{4}{3}$ .
  - To find $C$ , set $x = -1$ : $3(-1) + 2 = C(-1)(-1 - 2) \implies -1 = 3C \implies C = -\frac{1}{3}$ .
- Step 4: Integrate the terms: $\int \left( \frac{-1}{x} + \frac{4/3}{x - 2} - \frac{1/3}{x + 1} \right) dx$ .
- Final Solution: $-\ln|x| + \frac{4}{3}\ln|x - 2| - \frac{1}{3}\ln|x + 1| + C$ .

Complex Denominators: If you encounter irreducible quadratics like $x^2 + 25$ , the result will involve an $\arctan$ function if there is no $x$ in the numerator for a u-sub.
Double Angle Formulas: Remember that for $\sin^2(\theta)$ or $\cos^2(\theta)$ integrals:
- $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$
- $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$
Simplification Tip: If an integral yields $\sin(2\theta)$ , use the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ to return to functions compatible with your original reference triangle data.

Student Question: Will the test specify which method to use (e.g., "Use Trig Sub") or just ask us to solve it?
Response: Generally, it will just say "Solve this." However, there may be a couple of questions where a specific method like integration by parts or trig sub is requested to verify that the student knows that specific technique.