Calculus 2 2026 05 27

Concepts of Improper Integrals

• Defining Improper Integrals:

  • An integral is considered improper if the bounds are infinite ($\pm \infty$) or if there is a discontinuity (vertical asymptote) at the upper bound, lower bound, or at any point within the interval of integration.

• Example Analysis: Discontinuity Within Bounds:

  • For the integral $\int_1^6 \frac{8}{\sqrt[7]{x-2}^8} \, dx$, the function is defined at the bounds $1$ and $6$. However, a discontinuity occurs at $x = 2$ because the denominator becomes zero.

  • To solve this, the integral must be split into two parts: $\int_1^2 \dots \, dx + \int_2^6 \dots \, dx$.

• Convergence and Divergence:

  • If an integral results in an undefined value or infinity after applying limits, it is said to diverge (does not converge).

  • If the result is a specific finite number, the integral converges.

• Example Analysis: Infinite Bounds (Upper Bound):

  • Consider $\int_2^{\infty} \frac{3}{4 \sqrt[3]{x+3}^4} \, dx$.

  • The function is rewritten using negative exponents: $\frac{3}{4}(x+3)^{-\frac{4}{3}}$.

  • The upper bound is replaced with a limit: $\lim_{t \to \infty} \int_2^t \dots \, dx$.

  • Integration yields: $\frac{3}{4} \times \frac{(x+3)^{-\frac{1}{3}}}{-\frac{1}{3}} = -\frac{9}{4}(x+3)^{-\frac{1}{3}}$.

  • Evaluating the limit $\lim_{t \to \infty} \left[ -\frac{9}{4} \times \frac{1}{\sqrt[3]{t+3}} \right] = 0$.

  • The final value is determined by the lower bound evaluation, leading to convergence.

• Example Analysis: Arctangent and Infinite Limits:

  • $\int_0^{\infty} \frac{1}{x^2+1} \, dx = [\arctan(x)]_0^{\infty}$.

  • $\lim_{t \to \infty} \arctan(t) = \frac{\pi}{2}$.

  • $\arctan(0) = 0$.

  • Result: $\frac{\pi}{2}$.

Advanced Integration Techniques and Examples

• Integration by Parts (Logarithmic and Algebraic):

  • Problem: $\int_1^e 7t^2 \ln(t) \, dt$.

  • Let $u = \ln(t)$ and $dv = 7t^2 \, dt$.

  • Then $du = \frac{1}{t} \, dt$ and $v = \frac{7}{3}t^3$.

  • Formula: $uv - \int v \, du$.

  • Evaluate: $\left[ \ln(t) \cdot \frac{7}{3}t^3 \right]_1^e - \int_1^e \frac{7}{3}t^3 \cdot \frac{1}{t} \, dt$.

  • Simplifies to: $\frac{7}{3}e^3 - [\frac{7}{9}t^3]_1^e = \frac{7}{3}e^3 - \frac{7}{9}e^3 + \frac{7}{9}$.

  • Common denominator: $\frac{21}{9}e^3 - \frac{7}{9}e^3 + \frac{7}{9} = \frac{14}{9}e^3 + \frac{7}{9}$.

• Integration by Parts (Trigonometric):

  • Problem: $\int (x+3) \sin(2x) \, dx$.

  • Let $u = x+3$ (to make it "go away" through reduction) and $dv = \sin(2x) \, dx$.

  • $du = dx$ and $v = -\frac{1}{2}\cos(2x)$.

  • Result: $\frac{-(x+3)\cos(2x)}{2} + \int \frac{1}{2}\cos(2x) \, dx = \frac{-(x+3)\cos(2x)}{2} + \frac{1}{4}\sin(2x) + C$.

• Trigonometric Substitution (Sine):

  • Problem: $\int \frac{x^2}{\sqrt{81-x^2}} \, dx$.

  • Substitution: $x = 9\sin(\theta)$, $dx = 9\cos(\theta) \, d\theta$, and $\sqrt{81-x^2} = 9\cos(\theta)$.

  • The integral becomes: $\int (9\sin(\theta))^2 \, d\theta = \int 81\sin^2(\theta) \, d\theta$.

  • Using the identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$.

  • Integration leads to $\frac{81}{2}(\theta - \frac{1}{2}\sin(2\theta)) + C$.

  • Back-substitute using the triangle: $\theta = \arcsin(\frac{x}{9})$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.

• Partial Fraction Decomposition (Linear Factors):

  • Problem: $\int \frac{7}{49-x^2} \, dx$.

  • Factor denominator: $(7-x)(7+x)$.

  • Setup: $\frac{7}{(7-x)(7+x)} = \frac{A}{7-x} + \frac{B}{7+x}$.

  • Multiply by denominator: $7 = A(7+x) + B(7-x)$.

  • Solving for constants:

    • If $x = 7$, $7 = 14A \implies A = \frac{1}{2}$.

    • If $x = -7$, $7 = 14B \implies B = \frac{1}{2}$.

  • Integral: $\frac{1}{2} \int \frac{1}{7-x} \, dx + \frac{1}{2} \int \frac{1}{7+x} \, dx = -\frac{1}{2}\ln|7-x| + \frac{1}{2}\ln|7+x| + C$.

• Partial Fraction Decomposition (Rational Function):

  • Problem: $\int \frac{5x+3}{(x-3)(x+2)} \, dx$.

  • Setup: $\frac{5x+3}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$.

  • Solving:

    • Set $x = 3$: $5(3)+3 = A(3+2) \implies 18 = 5A \implies A = \frac{18}{5}$.

    • Set $x = -2$: $5(-2)+3 = B(-2-3) \implies -7 = -5B \implies B = \frac{7}{5}$.

  • Result: $\frac{18}{5}\ln|x-3| + \frac{7}{5}\ln|x+2| + C$.

Trigonometric Integration and Limits

• Powers of Trigonometric Functions:

  • For $\int \cos^3(x) \sin(x) \, dx$, use $u = \cos(x)$ because $\sin(x)$ is the $-du$.

  • For $\int \sin^2(3x) \, dx$, use the double angle formula: $\frac{1}{2}(1 - \cos(6x))$.

  • Integration results in $\frac{1}{2}x - \frac{1}{12}\sin(6x) + C$.

• Important Trigonometric Values for Integration:

  • $\arctan(\sqrt{3}) = \frac{\pi}{3}$.

  • $\arctan(1) = \frac{\pi}{4}$.

  • $\arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.

  • Vertical asymptotes of $\tan(x)$ at $x = \pm \frac{\pi}{2}$ translate to horizontal asymptotes for $\arctan(x)$.

  • $\lim_{t \to -\infty} \arctan(t) = -\frac{\pi}{2}$.

Questions & Discussion

• Question: When was the last time you actually used regular log (not natural log)?

• Answer: Likely while teaching. The speaker notes that natural log is the standard in high-level calculus and engineering, while regular log is less frequently used.

• Question: What does $y$ signify in the derivative expression $y \cdot \ln(x) + 1$?

• Answer: The speaker was just "goofing around" with an interesting derivative; it wasn't specifically tied to a class problem.

• Question: What is the limit of $\arctan(t)$ as $t \to -\infty$?

• Answer: It is $-\frac{\pi}{2}$, derived from the vertical asymptotes of the tangent function.

• Question: Can we use AI like ChatGPT for calculus homework?

• Answer: AI is a predictive language model. It may give results that look correct but are not actually computable because calculus problems are often "carefully crafted" and most functions do not have elementary anti-derivatives. If the AI is off by one small detail, the entire problem changes.