Calculus II: Integration Techniques, Applications of the Integral, and Parametric Equations Notes

Foundational Algebra Review and Common Errors

• Squaring a Binomial: A recurring issue on examinations is the incorrect expansion of a squared binomial. Students often mistakenly assume that $(a + b)^2$ equals $a^2 + b^2$ or $(a - b)^2$ equals $a^2 - b^2$.

• Correct Expansion Methods:

  • The Shortcut: To square a binomial, square the first term, take the product of the two terms and double it, then add the square of the last term.

    • Formula 1: $(a + b)^2 = a^2 + 2ab + b^2$

    • Formula 2: $(a - b)^2 = a^2 - 2ab + b^2$

  • FOIL Method: In the worst-case scenario, one should write the expression out as $(a - b) \times (a - b)$ and use the FOIL (First, Outer, Inner, Last) method to ensure accuracy.

Trigonometric Substitution Framework

• Purpose: Trigonometric substitution is used when an integrand contains a radical involving two terms (typically a variable squared and a constant squared) to transform it into a single-term expression within the radical.

• Core Identities: The technique relies entirely on the Pythagorean theorem and related identities:

  • $1 - \sin^2(\theta) = \cos^2(\theta)$

  • $1 + \tan^2(\theta) = \sec^2(\theta)$

  • $\sec^2(\theta) - 1 = \tan^2(\theta)$

• Substitution Selection Criteria:

  • Sine Substitution: Use $x = a \sin(\theta)$ for forms like $\sqrt{a^2 - x^2}$ (Constant squared minus variable squared).

  • Tangent Substitution: Use $x = a \tan(\theta)$ for forms like $\sqrt{a^2 + x^2}$ (Constant squared plus variable squared).

  • Secant Substitution: Use $x = a \sec(\theta)$ for forms like $\sqrt{x^2 - a^2}$ (Variable squared minus constant squared).

Detailed Integration Example: Sine Substitution

• Problem Statement: Evaluate the integral of $\frac{x^2}{\sqrt{36 - x^2}}$ from $x = 0$ to $x = 3$.

• Step 1: Identifying the Substitution: Since the form is $\sqrt{36 - x^2}$, where $a^2 = 36$, we let $x = 6 \sin(\theta)$.

• Step 2: Differential and Trig Transformation:

  • $dx = 6 \cos(\theta) d\theta$

  • The radical portion becomes $\sqrt{36 - 36 \sin^2(\theta)} = \sqrt{36(1 - \sin^2(\theta))} = \sqrt{36 \cos^2(\theta)} = 6 \cos(\theta)$.

• Step 3: Changing the Limits of Integration:

  • Upper limit: When $x = 3$, then $3 = 6 \sin(\theta)$, which means $\frac{1}{2} = \sin(\theta)$. On the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\theta = \frac{\pi}{6}$.

  • Lower limit: When $x = 0$, then $0 = 6 \sin(\theta)$, so $\theta = 0$.

• Step 4: Simplifying the Integral:

  • The integral becomes $\int_0^{\frac{\pi}{6}} \frac{36 \sin^2(\theta)}{6 \cos(\theta)} \times 6 \cos(\theta) d\theta$.

  • After canceling $6 \cos(\theta)$, we are left with $36 \int_0^{\frac{\pi}{6}} \sin^2(\theta) d\theta$.

• Step 5: Applying Reduction Identity: Use the half-angle identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$.

  • The integral becomes $18 \int_0^{\frac{\pi}{6}} (1 - \cos(2\theta)) d\theta = 18 [ \theta - \frac{\sin(2\theta)}{2} ]_0^{\frac{\pi}{6}}$.

• Step 6: Final Evaluation:

  • $18 [ \frac{\pi}{6} - \frac{\sin(\frac{\pi}{3})}{2} ] = 18 [ \frac{\pi}{6} - \frac{\sqrt{3}}{4} ] = 3\pi - \frac{9\sqrt{3}}{2}$.

Product-to-Sum Trigonometric Identities

• Application: Useful for integrals where the arguments (angles) of trigonometric functions are different.

• Case Study (Test Problem 8): Evaluate $\int 7x \sin(7x) \cos(3x) dx$.

  • Identity Provided: $\sin(A) \cos(B) = \frac{1}{2}[\sin(A - B) + \sin(A + B)]$.

  • Applying the identity where $A = 7x$ and $B = 3x$:

    • $\frac{1}{2} [ \sin(4x) + \sin(10x) ]$

  • The integral becomes $\frac{1}{2} \int (\sin(4x) + \sin(10x)) dx$.

  • Final Integration: $\frac{1}{2} [ \frac{-\cos(4x)}{4} - \frac{\cos(10x)}{10} ] + C$.

Integration by Parts Techniques

• Case Study (Test Problem 3): Evaluate $\int \theta \sec(\theta) \tan(\theta) d\theta$.

  • Step 1: Selection of u and dv: Let $u = \theta$ (simplifies upon differentiation) and $dv = \sec(\theta) \tan(\theta) d\theta$.

  • Step 2: Differentiating and Integrating: $du = d\theta$ and $v = \sec(\theta)$.

  • Step 3: Integration by Parts Formula ($\int u dv = uv - \int v du$):

    • $\theta \sec(\theta) - \int \sec(\theta) d\theta$

  • Final Step: The integral of $\sec(\theta)$ is $\ln|\sec(\theta) + \tan(\theta)|$.

    • Solution: $\theta \sec(\theta) - \ln|\sec(\theta) + \tan(\theta)| + C$.

Secant Substitution Nuances

• Case Study (Test Problem 10): Evaluate $\int \frac{1}{\sqrt{9x^2 - 1}} dx$.

  • Method 1: Direct Mapping: Recognize $9x^2$ as $(3x)^2$. Let $3x = \sec(\theta)$, then $3 dx = \sec(\theta) \tan(\theta) d\theta$, so $dx = \frac{1}{3} \sec(\theta) \tan(\theta) d\theta$.

  • Method 2: Factoring the Constant: Alternatively, factor out a 9 from the radical: $\sqrt{9(x^2 - \frac{1}{9})} = 3\sqrt{x^2 - (\frac{1}{3})^2}$. This leads to the substitution $x = \frac{1}{3} \sec(\theta)$.

  • Simplification: The radical $\sqrt{\sec^2(\theta) - 1}$ becomes $\tan(\theta)$.

  • Integration: The integral reduces to $\frac{1}{3} \int \sec(\theta) d\theta = \frac{1}{3} \ln|\sec(\theta) + \tan(\theta)| + C$.

  • Reversing Substitution via Triangle: If $\sec(\theta) = 3x$, then the hypotenuse is $3x$ and the adjacent side is 1. The opposite side is $\sqrt{9x^2 - 1}$. Therefore, $\tan(\theta) = \sqrt{9x^2 - 1}$.

  • Solution: $\frac{1}{3} \ln|3x + \sqrt{9x^2 - 1}| + C$.

Arc Length (Section 8.1)

• General Formula: $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$.

• Detailed Example (Problem 21): Find the length of $y = \frac{1}{4}x^2 - \frac{1}{2}\ln(x)$ from $x=1$ to $x=2$.

  • Finding the Derivative ($y'$): $y' = \frac{1}{2}x - \frac{1}{2x}$.

  • Simplifying $(y')^2$: Writing $y'$ as a single fraction $\frac{x^2 - 1}{2x}$ is often easier for squaring.

  • Radical Simplification: The expression $1 + (\frac{x^2 - 1}{2x})^2$ simplifies to $\frac{4x^2 + x^4 - 2x^2 + 1}{4x^2} = \frac{x^4 + 2x^2 + 1}{4x^2}$.

  • Perfect Square Trick: Note that $x^4 + 2x^2 + 1 = (x^2 + 1)^2$.

  • Taking the square root: $\frac{x^2 + 1}{2x} = \frac{x}{2} + \frac{1}{2x}$.

  • Integral: $L = \int_1^2 (\frac{x}{2} + \frac{1}{2x}) dx = [\frac{x^2}{4} + \frac{1}{2}\ln(x)]_1^2 = (1 + \frac{1}{2}\ln(2)) - (\frac{1}{4} + 0) = \frac{3}{4} + \frac{1}{2}\ln(2)$.

Surface Area (Section 8.2)

• General Formula (Rotation about x-axis): $S = \int_a^b 2\pi y \sqrt{1 + (\frac{dy}{dx})^2} dx$.

• Example Case (Problem 11): Surface area generated by rotating $y^2 = x + 1$ about the x-axis for $0 \leq x \leq 3$.

  • Choosing the top half of the parabola: $y = \sqrt{x + 1}$.

  • Derivative: $y' = \frac{1}{2\sqrt{x + 1}}$.

  • Radical: $\sqrt{1 + \frac{1}{4(x+1)}} = \sqrt{\frac{4x + 4 + 1}{4(x + 1)}} = \frac{\sqrt{4x + 5}}{2\sqrt{x + 1}}$.

  • Integral Setup: $S = \int_0^3 2\pi \sqrt{x + 1} \frac{\sqrt{4x + 5}}{2\sqrt{x + 1}} dx$.

  • Cancellation: The terms $\sqrt{x + 1}$ and 2 cancel out, leaving $\pi \int_0^3 \sqrt{4x + 5} dx$.

  • U-Substitution: Let $u = 4x + 5$, $du = 4 dx$. Limits: $x=0 \rightarrow u=5$, $x=3 \rightarrow u=17$.

  • Result: $\pi [\frac{2}{3} \times \frac{1}{4} u^{\frac{3}{2}}] = \frac{\pi}{6} [17^{\frac{3}{2}} - 5^{\frac{3}{2}}]$.

Parametric Equations (Section 10.1)

• Definitions: X and Y are given as functions of a parameter $t$ (often representing time).

• Graphing by Hand:

  • Create a table of values for $t$, calculate corresponding $x(t)$ and $y(t)$, and plot as coordinates $(x, y)$.

  • Indicate the direction of motion with arrows on the curve as $t$ increases.

• Calculator Procedures (TI-84):

  • Switch Mode from Function to Parametric (PAR).

  • Y= will now show X1T and Y1T inputs.

  • Window settings require Tmin, Tmax, and Tstep. A smaller Tstep results in a smoother curve but slower processing.

  • Zoom Square (Zoom 5) is essential to ensure circles look like circles and are not distorted by the screen aspect ratio.

• Eliminating the Parameter:

  • Solve for $t$ in one equation and substitute into the other, or use identities to link the variables.

  • Circles and Ellipses: Equations involving sine and cosine typically relate via $(\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = 1$.

  • If $a=b$, it is a circle. If $a \neq b$, it is an ellipse.

Questions & Discussion

• Student Question: Why did the teacher move from 36 to 18 on the outside of the integral?

• Instructor Response: This occurred during the application of the power-reduction identity, where $\sin^2(\theta)$ was replaced by $\frac{1}{2}(1 - \cos(2\theta))$. The coefficient 36 multiplied by the $\frac{1}{2}$ from the identity resulted in 18.

• Student Question: What is the definition of a doctor who gets all C's in medical school?

• Instructor Response: A doctor. The point is that Calculus 2 is difficult, and perseverance is key.

• Student Question: Is the homework the same as the study guide?

• Instructor Response: Yes. The homework is the study guide. All types of problems appearing on the test are represented in the homework assignments.

• Student Question: Do we need to worry about negative $t$ values in the real world?

• Instructor Response: In a mathematical context, $t$ is just a parameter. On a graph, $t=0$ is simply an arbitrary starting point. Negative values effectively represent the state of the system before that arbitrary start point.