Calculus II Study Guide: Advanced Integration and Improper Integrals

Advanced Techniques for Integrals with Quadratic Denominators

• Initial Evaluation of the Denominator

  • When faced with an integral like $\int_0^1 \frac{x}{x^2+4x+13} dx$, standard u-substitution fails if the numerator is not a multiple of the denominator's derivative.
  • Factoring is the first preference; however, in the expression $x^2+4x+13$, no two factors of $13$ add up to $4$. Since it cannot be factored as a binomial, it is classified as an irreducible quadratic.

• The Completing the Square Method

  • To integrate irreducible quadratics, use completing the square on the x-terms to transform the denominator into two terms (a squared binomial and a constant).
  • Process for $x^2+4x+13$:
    1. The coefficient of $x^2$ must be $1$.
    2. Take half the coefficient of $x$ ($\frac{4}{2} = 2$) and square it ($2^2 = 4$).
    3. Add and subtract this value within the expression: $(x^2 + 4x + 4) - 4 + 13$.
    4. Group and simplify: $(x+2)^2 + 9$.

• Simplification via U-Substitution

  • Let $u = x + 2$, then $du = dx$.
  • Changing Limits:
    • If $x = 1$, then $u = 1 + 2 = 3$.
    • If $x = 0$, then $u = 0 + 2 = 2$.
  • Rewriting the Integrand:
    • Since $u = x + 2$, algebra dictates that $x = u - 2$.
    • The new integral is: $\int_2^3 \frac{u-2}{u^2+9} du$.

• The "Divide and Conquer" Problem-Solving Technique

  • A complicated problem is broken down into two or more simpler problems. Solving the simpler parts solves the whole.
  • The integral is split at the numerator: $\int_2^3 \frac{u}{u^2+9} du - \int_2^3 \frac{2}{u^2+9} du$.

• Evaluating Split Integrals

  • Leftmost Integral (Substitution):
    • Let $t = u^2 + 9$, then $dt = 2u du$, or $\frac{1}{2} dt = u du$.
    • New limits for $t$: If $u=2, t=13$; if $u=3, t=18$.
    • Integrand: $\frac{1}{2} \int_{13}^{18} \frac{1}{t} dt = \frac{1}{2} [\ln|t|]_{13}^{18} = \frac{1}{2} (\ln(18) - \ln(13))$.
  • Rightmost Integral (Inverse Tangent Formula):
    • Formula: $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$.
    • Here, $a^2 = 9$, so $a = 3$.
    • Evaluating: $2 \times [\frac{1}{3} \arctan(\frac{u}{3})]_2^3 = \frac{2}{3} (\arctan(1) - \arctan(\frac{2}{3}))$.
    • Since $\arctan(1) = \frac{\pi}{4}$, this simplifies to $\frac{2}{3} (\frac{\pi}{4} - \arctan(\frac{2}{3}))$.

• Final Result and Arithmetic

  • Exact answer: $\frac{1}{2} \ln(\frac{18}{13}) - \frac{\pi}{6} + \frac{2}{3} \arctan(\frac{2}{3})$.
  • Note: Logarithm rules allow $\frac{1}{2} \ln(\frac{18}{13})$ to be written as $\ln(\sqrt{\frac{18}{13}})$, though decimal approximation using the former is often easier in calculators.

Partial Fraction Decomposition and Irreducible Quadratics

• Defining Irreducible Quadratics

  • A quadratic is irreducible in real numbers if it has no real zeros. Zeros can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
  • For $x^2 + 4x + 13 = 0$:
    • $x = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2}$.
    • $x = \frac{-4 \pm 6i}{2} = -2 \pm 3i$.
  • While technically factorable as $(x - (-2 + 3i))(x - (-2 - 3i))$, avoiding complex numbers and imaginary parts is the primary reason we use irreducible quadratics and the inverse tangent formula instead of partial fractions.

• The Multiplicity Requirement

  • Partial Fraction Decomposition (PFD) requires at least two factors in the denominator to be useful. With only one irreducible factor like $u^2 + 9$, PFD would only return the original expression.

Integrating with Rationalizing Substitutions

• Case Study: Problem 51

  • Integral: $\int \frac{e^{2x}}{e^{2x} + 3e^x + 2} dx$.
  • Strategy: Make a substitution that expresses the integrand as a rational function (polynomial divided by polynomial).
  • Let $u = e^x$, then $du = e^x dx$.
  • Identity: $e^{2x} = (e^x)^2 = e^x \times e^x$.
  • Transformation: $\int \frac{e^x \times e^x}{(e^x)^2 + 3e^x + 2} dx = \int \frac{u}{u^2 + 3u + 2} du$.

• Partial Fraction Decomposition Steps

  1. Factor the Denominator: $u^2 + 3u + 2 = (u + 1)(u + 2)$.
  2. Set up the Template: $\frac{u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}$.
  3. Clear Fractions: $u = A(u + 2) + B(u + 1)$.
  4. Solve for Constants:
     • Let $u = -2$: $-2 = B(-1) \Rightarrow B = 2$.
     • Let $u = -1$: $-1 = A(1) \Rightarrow A = -1$.
  5. Integrate: $\int (\frac{-1}{u+1} + \frac{2}{u+2}) du = -\ln|u+1| + 2\ln|u+2| + C$.
  6. Back-substitute: $-\ln(e^x + 1) + 2\ln(e^x + 2) + C$.
  7. Note on Absolute Values: Since $e^x$ is always positive, both terms $(e^x + 1)$ and $(e^x + 2)$ are always positive, making absolute value bars optional.

• Case Study: Problem 53

  • Integral: $\int \frac{\sec^2(t)}{\tan^2(t) + 3\tan(t) + 2} dt$.
  • Substitution: Let $u = \tan(t)$, then $du = \sec^2(t) dt$.
  • This instantly yields the rational function $\int \frac{1}{u^2 + 3u + 2} du$, solvable via PFD.

General Integration Strategies (Section 7.5 Review)

• Definitions of Functions

  • Rational Function: A polynomial divided by another polynomial ($\frac{P(x)}{Q(x)}$).
  • Proper Rational Function: The degree of the numerator is lower than the degree of the denominator.
  • Polynomial: A sum of power functions with positive integer exponents (e.g., $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$). Functions like $\sec(t)$ are not polynomials.

• Analytical Techniques for Random Integrals

  • Problem 12: $\int \frac{\arcsin(x)}{\sqrt{1-x^2}} dx$. Use u-sub where $u = \arcsin(x)$ because its derivative is present.
  • Problem 14: $\int \frac{x}{(2x+1)^3} dx$. Use u-sub $u = 2x + 1$. Solve for $x = \frac{u-1}{2}$. This turns the denominator into a single term ($u^3$), allowing fractions to be split and reduced as power functions.
  • Problem 16: $\int x \sin(x) \cos(x) dx$. Use the double angle identity $\sin(2x) = 2 \sin(x) \cos(x)$ first, then apply Integration by Parts.
  • Problem 24: $\int \frac{x^2}{\sqrt{1-x^2}} dx$. Trig substitution is required ($x = \sin(\theta)$).
  • Problem 28: $\int \frac{1}{1+e^{2a}} e^x dx$. Identification of variables is key: since the variable of integration is $x$ (noted by $dx$), the term $\frac{1}{1+e^{2a}}$ is a constant and can be factored out.

Type 1 Improper Integrals: Infinite Limits

• Definition

  • An improper integral of Type 1 occurs when one or both limits of integration are infinite ($\infty$ or $-\infty$).
  • Crucial Rule: Infinity is not a number and cannot be treated as one in calculations. It must be handled using limits.

• Evaluating with Limits

  • Step 1: Replace $\infty$ with a variable (e.g., $p$ or $a$).
  • Step 2: Setup the limit: $\lim_{p \to \infty} \int_1^p f(x) dx$.
  • Step 3: Evaluate the definite integral.
  • Step 4: Take the limit of the result.

• Concept of Convergence vs. Divergence

  • Example 1: $\int_1^\infty \frac{1}{x^3} dx = \lim_{p \to \infty} [-\frac{1}{2x^2}]_1^p = \lim_{p \to \infty} (-\frac{1}{2p^2} + \frac{1}{2})$.
    • As $p \to \infty$, $\frac{1}{2p^2} \to 0$. The value is $\frac{1}{2}$. This integral converges.
  • Example 2: $\int_1^\infty \frac{1}{x} dx = \lim_{a \to \infty} [\ln(x)]_1^a = \lim_{a \to \infty} \ln(a) - 0$.
    • The graph of $\ln(x)$ rises slowly but indefinitely. The limit is $\infty$. This integral diverges (does not exist).

• The P-Integral Rule

  • The integral $\int_1^\infty \frac{1}{x^p} dx$ will converge if and only if $p > 1$.
  • If $p = 1$, the function does not approach zero fast enough for the area to be finite.

• Double Infinite Limits

  • For $\int_{-\infty}^{\infty} f(x) dx$, break the integral into two pieces: $\int_{-\infty}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx$.
  • Commonly, $a = 0$ is chosen for convenience.

Type 2 Improper Integrals: Discontinuities

• Definition

  • Type 2 improper integrals occur when the integrand is undefined at one of the limits or at a point within the interval (usually a vertical asymptote).

• Evaluation at a Boundary Undefined point

  • Example: $\int_0^5 \frac{1}{(5-x)^{1/3}} dx$.
  • The function is undefined at $x=5$. Replace $5$ with $p$ and take the limit as $p \to 5$ from the left ($p \to 5^-$).
  • Calculation:
    1. U-sub: $u = 5 - x, du = -dx$.
    2. New limits: if $x = 0, u = 5$; if $x = p, u = 5 - p$.
    3. Integrate: $- \int_5^{5-p} u^{-1/3} du = -[\frac{3}{2} u^{2/3}]_5^{5-p}$.
    4. Limit: $\lim_{p \to 5^-} -\frac{3}{2} ((5-p)^{2/3} - 5^{2/3})$.
    5. Using Direct Substitution: $\lim_{p \to 5^-} (5-p)^{2/3} = 0$.
    6. Final value: $\frac{3}{2} \times 5^{2/3}$.

• Evaluation with an Internal Discontinuity

  • Example: $\int_0^5 \frac{w}{w-2} dw$.
  • The integrand is undefined at $w=2$. You must split the integral at the asymptote:
    • $\int_0^2 \frac{w}{w-2} dw + \int_2^5 \frac{w}{w-2} dw$.
  • Express both as limits: $\lim_{t \to 2^-} \int_0^t \dots + \lim_{n \to 2^+} \int_n^5 \dots$.

Questions & Discussion

• Question: Could one take the one-half into the logarithm like a square root?

  • Response: Yes, using the power rule for logs ($\frac{1}{2} \ln(18/13) = \ln(\sqrt{18/13})$), but it offers no significant advantage for calculation.

• Question: Does our result for $1/x^p$ remind us of hospital's rule?

  • Response: L'Hôpital's rule is used for comparing rates of growth or decay. Here, we are looking at whether a sum (area) remains bounded as we integrate toward problematic points. If it is unbounded, it diverges; if it has a horizontal asymptote in its sum-value, it converges.

• Question: If the asymptote is in the middle of the interval, do you approach from the left and right?

  • Response: Exactly. You break it into two separate intervals, one approaching the value from the left and one from the right.