Calculus II: Trig Substitutions, Completing the Square, and Partial Fraction Decomposition, and Exam Prep

Symmetry and Integrals of Odd and Even Functions

• Odd Functions and Symmetry:
  • For an odd function, such as $\sin(x)$, the integral with symmetric limits about zero (e.g., $\int_{-a}^{a} \sin(x)\,dx$) results in $0$.
  • Explanation: The area on the negative side (below the x-axis) is exactly offset by the area on the positive side (above the x-axis).
• Even Functions and Symmetry:
  • For an even function, such as $\sin^2(x)$ (represented as a black graph in the instructor's visual), the integral over symmetric limits can be simplified.
  • Formula: $\int_{-a}^{a} \sin^2(x)\,dx = 2 \times \int_{0}^{a} \sin^2(x)\,dx$.

Test Two Logistics and Preparation

• Schedule and Scope:
  • Date: Next Monday.
  • Coverage: Sections 6.2, 6.3, 6.5, 7.1, 7.2, and 7.3.
  • Extra Credit: A potential extra credit problem may involve Section 7.4.
• Duration and Timing:
  • Total time: 2 hours and 20 minutes ($140$ minutes).
  • Problem Estimate: Based on a rate of approximately $10$ minutes per problem, the test is expected to have around $14$ questions.
• Test Structure (Two-Part Format):
  • Part 1: Non-Calculator / No Note Portion: Students must put away all aids for this section.
  • Part 2: Open Calculator / Open Note Portion: Students may use a single page of notes (front and back) and a calculator.
  • Workflow Strategy: The instructor starts students with Part 2. Students can turn it in unfinished to start Part 1. If time remains after finishing Part 1, they may return to Part 2.
  • Warning: Do not get stuck on time-consuming problems in Part 2, as it might prevent completion of the more straightforward problems in Part 1.
• Note Sheet Guidance:
  • The sheet should improve over the semester.
  • Include trigonometric identities and reminders based on previous mistakes from Test One.
• Grading and Point Values:
  • Total Points: Likely between $130$ and $150$ points (compared to $108$ on the previous test).
  • Partial Credit Policy: Never leave a problem blank. A score of $-10$ indicates a completely blank response. The instructor once gave a student $2$ points (a $-8$) just for writing "I have no idea how to do this," though this is not a "winning strategy" for passing.
  • Error Penalties:
    • Simple sign errors: Approximately $-1$ point.
    • "Bad Algebra" (e.g., squaring a binomial incorrectly): Significant point deductions to draw attention to the mistake.

Questions & Discussion: Exam and Laboratory Details

• Question: How many questions will be on the test?
• Response: Given $140$ total minutes and an average of $10$ minutes per problem, there will likely be around $14$ questions. Many will involve "setup" rather than full integration, as setup (especially for volume problems) is often the more difficult part.
• Question: Is there a study session?
• Response: Yes, Sunday from $6$ to $8$ PM via the regular Zoom link. The instructor noted that Spectrum internet can be "flaky" but he has a backup hub.
• Question: Can I look at my cheat sheet for a few minutes before starting the non-note part?
• Response: Yes. Write your name on the calculator part and go for it. The instructor advises against copying problems onto scratch paper and then transferring them, as it burns too much time.
• Question: How many points are taken off for silly algebra vs. conceptual understanding?
• Response: A sign error is typically $1$ point. "Bad algebra," like squaring a binomial incorrectly, results in a fair number of points taken off to emphasize that such errors can "kill you in this class."

Trigonometric Integrals (Section 7.3 Homework)

• Strategy for $\int \tan^m(x) \sec^n(x)\,dx$:
  • Goal: Use u-substitution by creating the necessary $du$ term.
  • If letting $u = \tan(x)$, you need a factor of $\sec^2(x)$ for $du$.
  • If letting $u = \sec(x)$, you need a factor of $\sec(x)\tan(x)$ for $du$.
• Example: $\int \tan^3(x) \sec(x)\,dx$ (Problem 27):
  • Step 1: Rewrite as $\int \tan^2(x) (\sec(x)\tan(x))\,dx$.
  • Step 2: Use identity $\tan^2(x) = \sec^2(x) - 1$.
  • Step 3: Substitute $u = \sec(x)$, making $du = \sec(x)\tan(x)\,dx$.
  • Step 4: The integral becomes $\int (u^2 - 1)\,du = \frac{1}{3}u^3 - u + C$.
  • Step 5: Back-substitute: $\frac{1}{3}\sec^3(x) - \sec(x) + C$.
• Sine and Cosine Strategy:
  • If the power of cosine is odd, save one cosine and change the rest to sines.
  • If the power of sine is odd, save one sine and change the rest to cosines.

Trigonometric Substitution Techniques

• Identifying Substitutions Based on Identities:
  • Form $\sqrt{a^2 - x^2}$: Use $x = a\sin(\theta)$ (Identity: $1 - \sin^2(\theta) = \cos^2(\theta)$).
  • Form $\sqrt{a^2 + x^2}$: Use $x = a\tan(\theta)$ (Identity: $1 + \tan^2(\theta) = \sec^2(\theta)$).
  • Form $\sqrt{x^2 - a^2}$: Use $x = a\sec(\theta)$ (Identity: $\sec^2(\theta) - 1 = \tan^2(\theta)$).
• Example: $\int \frac{\sqrt{x^2-1}}{x^4}\,dx$ (Problem 11):
  • Substitution: Let $x = \sec(\theta)$, so $dx = \sec(\theta)\tan(\theta)\,d\theta$.
  • Simplification: $\int \frac{\sqrt{\sec^2(\theta)-1}}{\sec^4(\theta)} \sec(\theta)\tan(\theta)\,d\theta = \int \frac{\tan(\theta)}{\sec^3(\theta)} \tan(\theta)\,d\theta = \int \frac{\tan^2(\theta)}{\sec^3(\theta)}\,d\theta$.
  • Conversion to Sine/Cosine: $\int \frac{\sin^2(\theta)}{\cos^2(\theta)} \times \cos^3(\theta)\,d\theta = \int \sin^2(\theta)\cos(\theta)\,d\theta$.
  • U-substitution: Let $u = \sin(\theta)$, $du = \cos(\theta)\,d\theta$. Integral is $\frac{1}{3}u^3 + C = \frac{1}{3}\sin^3(\theta) + C$.
  • Reverse Substitution (Triangle Method):
    • Since $\sec(\theta) = \frac{x}{1}$, the hypotenuse is $x$, adjacent side is $1$, and opposite side is $\sqrt{x^2-1}$.
    • Result: $\frac{1}{3} (\frac{\sqrt{x^2-1}}{x})^3 + C = \frac{(x^2-1)^{3/2}}{3x^3} + C$.
• Example: $\int_{0}^{a} x^2\sqrt{a^2-x^2}\,dx$ (Problem 21):
  • Substitution: Let $x = a\sin(\theta)$, $dx = a\cos(\theta)\,d\theta$.
  • Changing Limits:
    • If $x=0 \rightarrow 0 = a\sin(\theta) \rightarrow \theta = 0$.
    • If $x=a \rightarrow a = a\sin(\theta) \rightarrow \sin(\theta) = 1 \rightarrow \theta = \frac{\pi}{2}$.
  • Simplification: $\int_{0}^{\pi/2} a^2\sin^2(\theta) \sqrt{a^2(1-\sin^2(\theta))} a\cos(\theta)\,d\theta = a^4\int_{0}^{\pi/2} \sin^2(\theta)\cos^2(\theta)\,d\theta$.
  • Using Identities: Uses the half-angle/double-angle identity $\sin^2(\theta) = \frac{1}{2}(1-\cos(2\theta))$ and $\cos^2(\theta) = \frac{1}{2}(1+\cos(2\theta))$.
  • Alternative Product Identity: $\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)$, lead to $\frac{a^4}{4}\int \sin^2(2\theta)\,d\theta$.
  • Final Evaluation: Applying the power reduction identity again results in $\frac{a^4\pi}{16}$ (assuming $a > 0$).

Advanced Integration: Completing the Square

• Concept: Used when the integrand contains an irreducible trinomial (quadratic). To apply trigonometric substitution, the trinomial must be converted to a binomial form.
• Example: $\int \frac{x^2}{(3 + 4x - 4x^2)^{3/2}}\,dx$ (Problem 32):
  • The process for $3 + 4x - 4x^2$:
    1. Factor out the coefficient of $x^2$ from the x-terms: $-4(x^2 - x) + 3$.
    2. Take half the linear coefficient ($-1$), square it ($1/4$), and add/subtract it: $-4(x^2 - x + \frac{1}{4} - \frac{1}{4}) + 3$.
    3. Factor the binomial: $-4[(x - \frac{1}{2})^2 - \frac{1}{4}] + 3$.
    4. Distribute: $-4(x - \frac{1}{2})^2 + 1 + 3 = 4 - 4(x - \frac{1}{2})^2$.
  • Substitution: Let $u = x - \frac{1}{2}$, which brings the integral to a standard trig sub form $\sqrt{a^2 - u^2}$.
  • Note: Trinomials requiring completing the square will appearing on the following test, not this Monday's test.

Introduction to Partial Fraction Decomposition (Section 7.4)

• Objective: To break a complex rational function (polynomial divided by polynomial) into a sum or difference of simpler fractions that are easy to integrate.
• Condition: The fraction must be "proper," meaning the degree of the numerator must be strictly less than the degree of the denominator. If improper, perform long division first.
• Case 1: Distinct Linear Factors:
  • Example: $\frac{3x+26}{x^2+x-12}$.
  1. Factor denominator: $(x+4)(x-3)$.
  2. Set up decomposition: $\frac{3x+26}{(x+4)(x-3)} = \frac{A}{x+4} + \frac{B}{x-3}$.
  3. Clear fractions: $3x+26 = A(x-3) + B(x+4)$.
  4. Solve for constants by picking strategic $x$-values:
     • Let $x=3 \rightarrow 35 = 7B \rightarrow B=5$.
     • Let $x=-4 \rightarrow 14 = -7A \rightarrow A=-2$.
  5. Integrate: $\int (\frac{-2}{x+4} + \frac{5}{x-3})\,dx = -2\ln|x+4| + 5\ln|x-3| + C$.
• Summary of PFD Cases:
  • Case 1: Distinct linear factors (e.g., $x, x+2, x-3$).
  • Case 2: Repeated linear factors (e.g., $(x+1)^2$). The setup must include every power: $\frac{A}{x+1} + \frac{B}{(x+1)^2}$.
  • Case 3: Irreducible quadratic factors (e.g., $x^2+4$). The numerator must be linear: $\frac{Ax+B}{x^2+4}$.
  • Case 4: Repeated irreducible quadratics (very complex, rarely assigned).

Questions & Discussion: Partial Fractions

• Question: What if the degrees of the numerator and denominator are the same?
• Response: It is still an improper fraction. You must perform long division first until you have a proper fraction remaining.
• Question: Why do we need the $\frac{A}{x+1}$ term for a repeated factor $(x+1)^2$?
• Response: Because when adding fractions to reach the common denominator of $(x+1)^2$, the original components could have had either denominator. You must account for all possible powers in the decomposition to find the correct constants.