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Single Variable Calculus
Differentiation
01The Derivative as the Slope of the Tangent Line02Secant Lines and the Limit Definition of the Derivative03Differentiating 1/x Straight From the Definition04Tangents to 1/x and the Triangle of Area 205Newton and Leibniz Notation, and the Power RuleProblem set0/10Problem set 20/10MIT problem set0/3
01The Derivative as an Instantaneous Rate of Change02Rates of Change Without Time: Gradient and Sensitivity03Easy Limits, 0/0, and One-Sided Limits04Continuity, Jumps, and Removable Discontinuities05Infinite Discontinuities: 1/x and Its Derivative06Differentiable Implies ContinuousProblem set0/10Problem set 20/10
01Two Kinds of Formula, and the Derivative of Sine02The Derivative of Cosine, and Two Limits at Zero03Bow and Bowstring: Proving the Two Trig Limits04Why the Trig Derivatives Hold Only in Radians05A Geometric Proof: Sine as a Height on the Circle06Finishing the Proof, and the Product and Quotient RulesProblem set0/10Problem set 20/10MIT problem set0/1
01Proving the Product Rule: Change One Factor at a Time02The Quotient Rule, and the Power Rule for Negative Exponents03The Chain Rule: Differentiating a Function of a Function04Higher Derivatives: The Sine Cycle and Three Notations05The nth Derivative of xⁿ Is n Factorial, by InductionProblem set0/10Problem set 20/10MIT problem set0/2
01The Power Rule for Rational Exponents02The Slope of a Circle, Two Ways03Implicit Differentiation of a Quartic Curve04Inverse Functions and the Reflection Across y = x05Derivatives of the Arctangent and ArcsineProblem set0/10Problem set 20/10MIT problem set0/1
01The Derivative of aˣ and the Mystery Number M(a)02Defining e by Slope One, and the Derivative of ln x03M(a) = ln a: Differentiating aˣ and xˣ04Why (1 + 1/n)ⁿ Approaches eProblem set0/10Problem set 20/10MIT problem set0/1
Applications of Differentiation
01Linear Approximation: A Curve Is Its Tangent Line02Five Linear Approximations at Zero03Combining Approximations, from Algebra to GPS Clocks04Quadratic Approximation: The Best-Fit ParabolaProblem set0/10Problem set 20/10MIT problem set0/1
01Linear Approximation as a Proportionality Rule02Quadratic Approximation: Why the ½, and Products03Curve Sketching: The Sign of f′ and Critical Points04Curve Sketching: Ends, Concavity and InflectionProblem set0/10Problem set 20/10MIT problem set0/1
01Sketching Without Critical Points: (x+1)/(x+2)02A Five-Step Curve-Sketching Scheme, Applied to x / ln x03x / ln x: Inflection Point and Where Max and Min Can OccurProblem set0/10Problem set 20/10
01Two Squares From One Wire: The Critical Point Gives the Least Area02The Open Box of Fixed Volume: Least Material When x = 2y03Related Rates: The Police Radar and the Conical TankProblem set0/10Problem set 20/10MIT problem set0/4
01The Hanging Ring: Why the String Makes Equal Angles02Newton's Method: Tangent Lines That Find the Square Root of Five03Why Newton's Method Doubles Its Digits, and How It Fails04The Mean Value Theorem: Somewhere You Hit Your Average Speed05Positive Derivative Means Increasing: Proving eˣ Exceeds 1 + xProblem set0/10Problem set 20/10MIT problem set0/2
Integration
01Differentials: dy = f′(x) dx and the Cube Root of 64.102Antiderivatives Differ Only by a Constant03Integration by Substitution and Advanced GuessingProblem set0/10Problem set 20/10MIT problem set0/1
01Separation of Variables: dy/dx = −xy and the Bell Curve02General Solutions: Signs, Constants and Initial Conditions03Orthogonal Trajectories: Parabolas and Crossing EllipsesProblem set0/10Problem set 20/10MIT problem set0/1
01The Definite Integral: Area as a Limit of Rectangle Sums02Area Under y = x²: Squeezing the Sum of Squares03Sigma Notation and the Area Pattern: b, b²/2, b³/304Riemann Sums: the Definite Integral and a Year of LoansProblem set0/10Problem set 20/10MIT problem set0/1
01The Fundamental Theorem of Calculus: Area in One Line02Net Change and Signed Area: Integrating Velocity03Six Rules for Definite Integrals and Bounds on eᵇ04Substitution in Definite Integrals: Change the LimitsProblem set0/10Problem set 20/10MIT problem set0/1

Integration by Substitution and Advanced Guessing

How does choosing u as the messiest piece turn a hard integral into an easy one, and why can two different answers both be right?


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Your summary note

    1. 1

      The method of substitution

      Record the steps (set uuu to the messiest piece, find dududu, rewrite and integrate in uuu, change back to xxx) and carry ∫x3(x4+2)5 dx\int x^3(x^4+2)^5\,dx∫x3(x4+2)5dx through to 124(x4+2)6+c\frac{1}{24}(x^4+2)^6 + c241​(x4+2)6+c.

    2. 2

      Advanced guessing: guess, differentiate, fix the coefficient

      Write out the guess and its derivative check for ∫x dx1+x2\int \frac{x\,dx}{\sqrt{1+x^2}}∫1+x2​xdx​, ∫e6x dx\int e^{6x}\,dx∫e6xdx and ∫xe−x2 dx\int x e^{-x^2}\,dx∫xe−x2dx, showing the correcting factor 16\frac{1}{6}61​ and −12-\frac{1}{2}−21​ where they appear.

    3. 3

      Two answers for ∫sin⁡xcos⁡x dx\int \sin x\cos x\,dx∫sinxcosxdx

      Derive both 12sin⁡2x+c1\frac{1}{2}\sin^2 x + c_121​sin2x+c1​ and −12cos⁡2x+c2-\frac{1}{2}\cos^2 x + c_2−21​cos2x+c2​, subtract them using sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1sin2x+cos2x=1, and state that the families agree with c2−c1=12c_2 - c_1 = \frac{1}{2}c2​−c1​=21​.

    4. 4

      Substituting u=ln⁡xu = \ln xu=lnx in ∫dxxln⁡x\int \frac{dx}{x\ln x}∫xlnxdx​

      Carry out the substitution with du=dxxdu = \frac{dx}{x}du=xdx​, splitting the integrand into 1u\frac{1}{u}u1​ and dududu, and reach ln⁡(ln⁡x)+c\ln(\ln x) + cln(lnx)+c, noting the absolute-value form ln⁡∣ln⁡x∣+c\ln|\ln x| + cln∣lnx∣+c.

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