5.5 Integration by Substitution (u-Substitution)
This is the reverse of the chain rule. The goal is to find a part of the integrand, which we'll call $u$, whose derivative $du$ is also present (or can be formed with a constant multiple).
Key Formula
If $u = g(x)$, then $du = g'(x)dx$, and the integral transforms:
$\int f(g(x))g'(x)dx = \int f(u)du$Example from Review (#1)
Evaluate $\int\frac{(\ln x)^{4}}{x}dx.$
Step 1: Identify u. Notice that the derivative of $\ln x$ is $\frac{1}{x}$, which is in the integrand. This makes $\ln x$ a great choice for $u$.
Let $u = \ln x$
Step 2: Find du.
Then $du = \frac{1}{x}dx$
Step 3: Substitute. Replace $\ln x$ with $u$ and $\frac{1}{x}dx$ with $du$.
$\int u^4 du$
Step 4: Integrate with respect to u.
$\frac{u^5}{5} + C$
Step 5: Substitute back to x.
$\frac{(\ln x)^5}{5} + C$
6.1 Integration by Parts
This technique is used for integrating products of functions. It's derived from the product rule for differentiation.
Key Formula
$\int u \, dv = uv - \int v \, du$A helpful mnemonic for choosing $u$ is LIATE:
- Logarithmic functions (e.g., $\ln x$)
- Inverse trigonometric functions (e.g., $\arctan x$)
- Algebraic functions (e.g., $x^2, 3x$)
- Trigonometric functions (e.g., $\sin x, \cos x$)
- Exponential functions (e.g., $e^x$)
Choose the function that comes first in LIATE as your $u$.
Example from Review (#8)
Evaluate $\int te^{-3t}dt$.
Step 1: Choose u and dv. Following LIATE, 't' is Algebraic and 'e-3t' is Exponential. So, we choose:
Let $u = t$ and $dv = e^{-3t}dt$
Step 2: Find du and v.
$du = dt$
$v = \int e^{-3t}dt = -\frac{1}{3}e^{-3t}$
Step 3: Apply the formula.
$\int te^{-3t}dt = (t)(-\frac{1}{3}e^{-3t}) - \int (-\frac{1}{3}e^{-3t})dt$
$= -\frac{t}{3}e^{-3t} + \frac{1}{3}\int e^{-3t}dt$
Step 4: Solve the remaining integral.
$= -\frac{t}{3}e^{-3t} + \frac{1}{3}(-\frac{1}{3}e^{-3t}) + C$
$= -\frac{t}{3}e^{-3t} - \frac{1}{9}e^{-3t} + C$
6.2 Trigonometric Integrals
These integrals involve powers of trigonometric functions. The strategy depends on the powers of the functions.
Key Identities to Memorize
$\sin^2 x + \cos^2 x = 1$$\tan^2 x + 1 = \sec^2 x$
$\sin^2 x = \frac{1 - \cos(2x)}{2}$
$\cos^2 x = \frac{1 + \cos(2x)}{2}$
$\sin(2x) = 2\sin x \cos x$
Strategy for $\int \sin^m x \cos^n x \, dx$
- If power of $\cos x$ is odd ($n=2k+1$), save one $\cos x$ factor, convert the rest to sines using $\cos^2 x = 1 - \sin^2 x$, then use $u = \sin x$.
- If power of $\sin x$ is odd ($m=2k+1$), save one $\sin x$ factor, convert the rest to cosines using $\sin^2 x = 1 - \cos^2 x$, then use $u = \cos x$.
- If both powers are even, use the half-angle identities repeatedly.
Example from Review (#14)
Evaluate $\int \sin^6 x \cos^3 x \, dx$.
Strategy: Power of cosine is odd.
$\int \sin^6 x \cos^2 x \cos x \, dx$
$= \int \sin^6 x (1 - \sin^2 x) \cos x \, dx$
Let $u = \sin x$, so $du = \cos x \, dx$
$= \int u^6(1 - u^2) \, du = \int (u^6 - u^8) \, du$
$= \frac{u^7}{7} - \frac{u^9}{9} + C$
$= \frac{\sin^7 x}{7} - \frac{\sin^9 x}{9} + C$
6.3 Integration by Partial Fractions
Used to integrate rational functions $\frac{P(x)}{Q(x)}$. The first step is to decompose the rational function into simpler fractions.
Decomposition Rules
1. Distinct Linear Factors: $\frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$
2. Repeated Linear Factors: $\frac{P(x)}{(ax+b)^k} = \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_k}{(ax+b)^k}$
3. Irreducible Quadratic Factors: $\frac{P(x)}{ax^2+bx+c} = \frac{Ax+B}{ax^2+bx+c}$
Example from Review (#20)
Evaluate $\int\frac{1}{(x+6)(x-2)}dx$.
Step 1: Decompose.
$\frac{1}{(x+6)(x-2)} = \frac{A}{x+6} + \frac{B}{x-2}$
Step 2: Solve for A and B. Multiply by the common denominator:
$1 = A(x-2) + B(x+6)$
Let $x=2$: $1 = A(0) + B(8) \implies B = \frac{1}{8}$
Let $x=-6$: $1 = A(-8) + B(0) \implies A = -\frac{1}{8}$
Step 3: Integrate.
$\int (\frac{-1/8}{x+6} + \frac{1/8}{x-2}) dx$
$= -\frac{1}{8}\ln|x+6| + \frac{1}{8}\ln|x-2| + C = \frac{1}{8}\ln|\frac{x-2}{x+6}| + C$
6.5 Numerical Approximation
When an integral is difficult or impossible to solve analytically, we can approximate its value.
Trapezoidal Rule
$\int_a^b f(x)dx \approx \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)]$where $\Delta x = \frac{b-a}{n}$
Simpson's Rule (n must be even)
$\int_a^b f(x)dx \approx \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + \dots + 4f(x_{n-1}) + f(x_n)]$where $\Delta x = \frac{b-a}{n}$
Interactive Calculator
6.6 Improper Integrals
Integrals with infinite limits of integration or with an infinite discontinuity within the interval.
Types
Type 1 (Infinite Interval): $\int_a^\infty f(x)dx = \lim_{t\to\infty} \int_a^t f(x)dx$
Type 2 (Discontinuous Integrand): If $f(x)$ is discontinuous at $b$, then $\int_a^b f(x)dx = \lim_{t\to b^-} \int_a^t f(x)dx$
An integral converges if the limit exists and is finite. Otherwise, it diverges.
Example from Review (#32)
Determine if $\int_{e}^{\infty}\frac{79}{x(\ln x)^{3}}dx$ converges or diverges.
Step 1: Set up the limit.
$\lim_{t\to\infty} \int_e^t \frac{79}{x(\ln x)^3} dx$
Step 2: Solve the integral (use u-sub).
Let $u = \ln x$, $du = \frac{1}{x}dx$.
$\int \frac{79}{u^3} du = 79 \int u^{-3} du = 79 \frac{u^{-2}}{-2} = -\frac{79}{2u^2} = -\frac{79}{2(\ln x)^2}$
Step 3: Evaluate the limit.
$\lim_{t\to\infty} [-\frac{79}{2(\ln x)^2}]_e^t = \lim_{t\to\infty} (-\frac{79}{2(\ln t)^2} - (-\frac{79}{2(\ln e)^2}))$
$= \lim_{t\to\infty} (-\frac{79}{2(\ln t)^2} + \frac{79}{2(1)^2})$
As $t \to \infty$, $\ln t \to \infty$, so $\frac{79}{2(\ln t)^2} \to 0$.
$= 0 + \frac{79}{2} = \frac{79}{2}$
Conclusion: The integral converges to $\frac{79}{2}$.
7.1 Area Between Curves
If $f(x) \ge g(x)$ on $[a, b]$, the area is: $A = \int_a^b [f(x) - g(x)] dx$
If $f(y) \ge g(y)$ on $[c, d]$, the area is: $A = \int_c^d [f(y) - g(y)] dy$
Interactive Visualizer
Note: Intersection points are calculated automatically. Visualization range is based on these points.
7.2 & 7.3 Volumes of Revolution
Disk Method
Rotation about x-axis: $V = \int_a^b \pi [R(x)]^2 dx$
Rotation about y-axis: $V = \int_c^d \pi [R(y)]^2 dy$
Washer Method
Rotation about x-axis: $V = \int_a^b \pi ([R(x)]^2 - [r(x)]^2) dx$
Rotation about y-axis: $V = \int_c^d \pi ([R(y)]^2 - [r(y)]^2) dy$
Cylindrical Shells Method
Rotation about y-axis: $V = \int_a^b 2\pi x h(x) dx$
Rotation about x-axis: $V = \int_c^d 2\pi y h(y) dy$
Interactive visualizer coming soon!
7.4 Arc Length
For a function $y=f(x)$ from $x=a$ to $x=b$: $L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$
For a function $x=g(y)$ from $y=c$ to $y=d$: $L = \int_c^d \sqrt{1 + [g'(y)]^2} \, dy$
Example from Review (#52)
Find the length of $y=1+2x^{3/2}$ for $0 \le x \le 1$.
Step 1: Find the derivative.
$y' = \frac{dy}{dx} = 2 \cdot \frac{3}{2}x^{1/2} = 3\sqrt{x}$
Step 2: Square the derivative.
$(y')^2 = (3\sqrt{x})^2 = 9x$
Step 3: Set up the integral.
$L = \int_0^1 \sqrt{1 + 9x} \, dx$
Step 4: Solve the integral (u-sub).
Let $u = 1+9x$, $du = 9dx \implies dx = \frac{1}{9}du$.
Change bounds: $x=0 \to u=1$, $x=1 \to u=10$.
$L = \int_1^{10} \sqrt{u} \frac{1}{9}du = \frac{1}{9} [\frac{2}{3}u^{3/2}]_1^{10}$
$= \frac{2}{27}[10^{3/2} - 1^{3/2}] = \frac{2}{27}(10\sqrt{10} - 1)$
7.6 Work
Work done by a variable force is the integral of the force. For pumping liquid, we integrate the work done to lift thin horizontal slices of the liquid.
Work Pumping Liquid
$W = \int_a^b (\text{density}) \cdot g \cdot A(y) \cdot D(y) \, dy$
- $A(y)$: Area of a horizontal slice at height $y$.
- $D(y)$: Distance the slice at height $y$ must be lifted.
- $g$: acceleration due to gravity ($9.8 \, m/s^2$ or $32 \, ft/s^2$).
- Density of water: $1000 \, kg/m^3$ or $62.5 \, lb/ft^3$.
Example from Review (#55a)
Inverted cone, height 8m, radius 4m. Full of water. Pump all water out (no spout).
Setup: Place origin at the bottom vertex. The y-axis goes up.
Find radius of a slice: By similar triangles, $\frac{r}{y} = \frac{4}{8} \implies r = \frac{y}{2}$.
Area of a slice $A(y)$: $A(y) = \pi r^2 = \pi (\frac{y}{2})^2 = \frac{\pi y^2}{4}$.
Distance to lift $D(y)$: A slice at height $y$ must travel to the top (height 8). So, $D(y) = 8-y$.
Integral bounds: The water is from $y=0$ to $y=8$.
Work Integral:
$W = \int_0^8 (1000)(9.8) \pi (\frac{y^2}{4})(8-y) dy$
8.1 & 8.2 Sequence and Series Convergence
Key Tests
Test for Divergence: If $\lim_{n\to\infty} a_n \neq 0$ or the limit DNE, then the series $\sum a_n$ diverges.
Geometric Series: $\sum_{n=1}^\infty ar^{n-1}$ converges to $\frac{a}{1-r}$ if $|r| < 1$. It diverges if $|r| \ge 1$.
Ratio Test: Let $L = \lim_{n\to\infty} |\frac{a_{n+1}}{a_n}|$.
- If $L < 1$, the series is absolutely convergent.
- If $L > 1$, the series is divergent.
- If $L = 1$, the test is inconclusive.
Example from Review (#59)
Determine if $\sum_{n=1}^{\infty}\frac{(-7)^{n-1}}{9^{n}}$ converges or diverges.
Step 1: Rewrite as a geometric series.
$\sum_{n=1}^{\infty}\frac{(-7)^{n-1}}{9 \cdot 9^{n-1}} = \sum_{n=1}^{\infty} \frac{1}{9} (\frac{-7}{9})^{n-1}$
Step 2: Identify a and r.
This is a geometric series with $a = \frac{1}{9}$ and $r = -\frac{7}{9}$.
Step 3: Check for convergence.
$|r| = |-\frac{7}{9}| = \frac{7}{9} < 1$. So, the series converges.
Step 4: Find the sum.
$S = \frac{a}{1-r} = \frac{1/9}{1 - (-7/9)} = \frac{1/9}{16/9} = \frac{1}{16}$.
8.5-8.7 Power, Taylor, & Maclaurin Series
Definitions
Power Series (centered at a): $\sum_{n=0}^\infty c_n (x-a)^n$
Taylor Series (centered at a): $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$
Maclaurin Series (Taylor series at a=0): $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$
Important Maclaurin Series to Memorize
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \dots$
$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1 + x + x^2 + \dots$, for $|x|<1$
9.1-9.4 Parametric & Polar Equations
Calculus with Parametric Equations
Curve: $x=f(t), y=g(t)$
Slope: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Arc Length: $L = \int_\alpha^\beta \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$
Calculus with Polar Equations
Curve: $r=f(\theta)$
Area: $A = \int_a^b \frac{1}{2} [f(\theta)]^2 \, d\theta = \int_a^b \frac{1}{2} r^2 \, d\theta$
Arc Length: $L = \int_a^b \sqrt{r^2 + (\frac{dr}{d\theta})^2} \, d\theta$