Integration (Math III)

Senior High School Mathematics III: Complete Integration Study Notes

The integration you learned in Mathematics II primarily focused on polynomial functions (such as quadratic and cubic functions). In Mathematics III Integration, we will integrate all kinds of functions you have encountered so far—including trigonometric, exponential, logarithmic, rational, and irrational functions.
At first, you might feel overwhelmed and think, "There are so many formulas to memorize...!" But don't worry! The essence of integration lies in being the "reverse operation of differentiation" and "adding up tiny, fine slices." Once you master the basic mechanisms and learn how to identify patterns, solving integrals will feel just like solving fun puzzles. Let's take it step by step!

*Note: This chapter assumes and builds upon the knowledge of limits and derivatives learned in "Differential Calculus."

1. Fundamentals of Indefinite Integrals and Formulas for Various Functions

When the derivative of a function \(F(x)\) is \(f(x)\)—that is, when \(F'(x) = f(x)\)—we call \(F(x)\) an antiderivative (or primitive function) of \(f(x)\). Adding a constant of integration \(C\) to this antiderivative gives the indefinite integral, which is represented using the integral symbol \(\int\) as follows:

\(\int f(x) dx = F(x) + C\) (\(C\) is the constant of integration)

Indefinite Integral Formulas for Various Functions

・Integration of Power Functions \(x^\alpha\)
When \(\alpha \neq -1\): \(\int x^\alpha dx = \frac{1}{\alpha + 1} x^{\alpha + 1} + C\)
When \(\alpha = -1\): \(\int \frac{1}{x} dx = \log |x| + C\)

・Integration of Trigonometric Functions
\(\int \sin x dx = -\cos x + C\)
\(\int \cos x dx = \sin x + C\)
\(\int \frac{1}{\cos^2 x} dx = \tan x + C\)

・Integration of Exponential Functions
\(\int e^x dx = e^x + C\)
\(\int a^x dx = \frac{a^x}{\log a} + C\) (\(a > 0, a \neq 1\))

[Key Point] Watch Out for the Signs in the Formulas!

[Key Point]
Be extra careful with functions whose signs turn out opposite to their differentiation counterparts!
・Differentiating \(\sin x\) gives \(\cos x\), but integrating \(\sin x\) gives \(-\cos x\).
・Differentiating \(\cos x\) gives \(-\sin x\), but integrating \(\cos x\) gives \(\sin x\).
Whenever you feel unsure, make it a habit to check: "If I differentiate the right-hand side, do I get the integrand back?"

[Common Mistake] Forgetting the Absolute Value Signs

[Common Mistake]
A remarkably common error is writing \(\int \frac{1}{x} dx = \log x + C\).
To account for cases where \(x\) takes negative values, you must always include absolute value bars on the argument: \(\log |x| + C\).

2. Integration by Substitution

Integration by substitution is a technique based on reversing the chain rule of differentiation. By replacing part of an expression with another variable (such as \(t\)), we transform a complicated integrand into a simpler, standard formula.

(1) Linear Substitution: The Form \(ax + b = t\)

This is the most fundamental substitution emphasized in the syllabus.
If we set \(ax + b = t\) and differentiate both sides with respect to \(x\), we get \(a = \frac{dt}{dx}\), meaning we can substitute \(dx = \frac{1}{a} dt\).

Formula:
\(\int f(ax + b) dx = \frac{1}{a} F(ax + b) + C\) (where \(F'(x) = f(x), a \neq 0\))

Example: \(\int (3x - 2)^4 dx\)
Let \(3x - 2 = t\), which gives \(3 dx = dt\), or \(dx = \frac{1}{3} dt\).
\(\int t^4 \cdot \frac{1}{3} dt = \frac{1}{3} \cdot \frac{1}{5} t^5 + C = \frac{1}{15} (3x - 2)^5 + C\)

(2) Integrals of the Form \(\int \frac{g'(x)}{g(x)} dx = \log |g(x)| + C\)

Whenever you see a fraction where the numerator is the derivative of the denominator, you can immediately integrate it into a logarithmic form.
Example: \(\int \frac{2x}{x^2 + 1} dx\)
Because differentiating the denominator \(g(x) = x^2 + 1\) yields the numerator \(2x\), the result is instantly found to be \(\log (x^2 + 1) + C\) (the absolute value can be dropped since \(x^2 + 1 > 0\)).

(3) Definite Integrals by Substitution and Trigonometric Substitution: \(x = a \sin \theta\)

When performing substitution in definite integrals, it is crucial that you convert the limits of integration (lower and upper bounds).

Formula for Definite Integration by Substitution:
\(\int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(t) dt\)

In particular, for definite integrals containing \(\sqrt{a^2 - x^2}\), the standard strategy is to substitute \(x = a \sin \theta\).
By setting \(\left(-\frac{\pi}{2} \leqq \theta \leqq \frac{\pi}{2}\right)\), we have \(\sqrt{a^2 - x^2} = \sqrt{a^2(1 - \sin^2 \theta)} = \sqrt{a^2 \cos^2 \theta} = a \cos \theta\), which neatly eliminates the square root!

Example: \(\int_0^a \sqrt{a^2 - x^2} dx\) (\(a > 0\))
Let \(x = a \sin \theta\), then \(dx = a \cos \theta d\theta\).
As \(x\) goes from \(0 \to a\), \(\theta\) changes from \(0 \to \frac{\pi}{2}\).
\(\int_0^{\frac{\pi}{2}} (a \cos \theta) \cdot (a \cos \theta) d\theta = a^2 \int_0^{\frac{\pi}{2}} \cos^2 \theta d\theta\)
Using the half-angle formula \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\(= a^2 \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} d\theta = a^2 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} a^2\)

[Fun Fact] Definite Integrals and the Area of a Circle

[Fun Fact]
The integral \(\int_0^a \sqrt{a^2 - x^2} dx\) in the example above represents the area of the region bounded by the curve \(y = \sqrt{a^2 - x^2}\) (the upper half of a circle of radius \(a\)), the \(x\)-axis, the \(y\)-axis, and the line \(x = a\). Since this is simply a quadrant (\(\frac{1}{4}\) sector) of a circle with radius \(a\), we can geometrically see why the area is \(\frac{1}{4} \pi a^2\) without doing any integration at all!

3. Integration by Parts

Integration by parts is derived by integrating both sides of the product rule for differentiation, \(\{f(x)g(x)\}' = f'(x)g(x) + f(x)g'(x)\). It shows its true power when integrating functions that appear as products.

Integration by Parts Formulas

Indefinite Integrals:
\(\int f(x) g'(x) dx = f(x) g(x) - \int f'(x) g(x) dx\)

Definite Integrals:
\(\int_a^b f(x) g'(x) dx = \left[ f(x) g(x) \right]_a^b - \int_a^b f'(x) g(x) dx\)

[Key Point] Which Should Be \(f(x)\) (to differentiate) and Which Should Be \(g'(x)\) (to integrate)?

[Key Point]
The Senior High School Mathematics III curriculum mainly focuses on "simple functions that yield results after a single application."
The trick to integration by parts is choosing the function that simplifies upon differentiation as \(f(x)\), and the function that is easy to integrate as \(g'(x)\).
・Polynomial \(\times\) Exponential: \((x) \cdot (e^x) \implies\) choose \(x\) to differentiate, \(e^x\) to integrate.
・Polynomial \(\times\) Trigonometric: \((x) \cdot (\cos x) \implies\) choose \(x\) to differentiate, \(\cos x\) to integrate.
・Expressions involving logarithmic functions: Since \(\log x\) differentiates into the simpler \(\frac{1}{x}\), always choose it as the part to differentiate.

Classic Example 1: \(\int x e^x dx\)
Set \(f(x) = x\) and \(g'(x) = e^x\), which gives \(f'(x) = 1\) and \(g(x) = e^x\).
\(\int x e^x dx = x e^x - \int 1 \cdot e^x dx = x e^x - e^x + C = (x - 1)e^x + C\)

Classic Example 2: \(\int \log x dx\)
There doesn't seem to be a product here, but viewing it as \(1 \cdot \log x\) is a famous, essential technique!
Set \(f(x) = \log x\) and \(g'(x) = 1\), giving \(f'(x) = \frac{1}{x}\) and \(g(x) = x\).
\(\int \log x dx = (\log x) \cdot x - \int \frac{1}{x} \cdot x dx = x \log x - \int 1 dx = x \log x - x + C\)

4. Applications of Definite Integrals (Area, Volume, and Arc Length)

By applying definite integrals, we can calculate not only plane areas but also volumes of solid figures and the lengths of curves.

(1) Area Between Curves

When \(f(x) \geqq g(x)\) across the interval \(a \leqq x \leqq b\), the area \(S\) of the region enclosed by the two curves \(y = f(x)\), \(y = g(x)\) and the two lines \(x = a\), \(x = b\) is given by:

\(S = \int_a^b \{ f(x) - g(x) \} dx\)

・If the curves cross and switch which one is on top, divide the interval at the intersection point and always integrate "(upper curve) \(-\) (lower curve)".
・If a curve is expressed in the form \(x = g(y)\) and it is simpler to integrate along the \(y\)-axis, compute it as \(S = \int_c^d g(y) dy\).

(2) Volume of Solids

When a solid is sliced by planes perpendicular to a given axis (usually the \(x\)-axis), letting \(S(x)\) be the cross-sectional area at coordinate \(x\), the volume \(V\) of the solid from \(x = a\) to \(x = b\) is obtained by continuously accumulating these cross sections.

Volume from Cross-Sectional Area:
\(V = \int_a^b S(x) dx\)

Volume of a Solid of Revolution (Rotation about the \(x\)-axis):
When the region bounded by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = a, x = b\) is revolved once about the \(x\)-axis, every cross section forms a circle of radius \(|y|\). Thus, the cross-sectional area is \(S(x) = \pi y^2 = \pi \{f(x)\}^2\).
Therefore, the volume formula is:

\(V = \pi \int_a^b y^2 dx = \pi \int_a^b \{f(x)\}^2 dx\)

(3) Arc Length of a Curve

The arc length across an infinitesimal interval can be seen as the hypotenuse of a right-angled triangle, \(\sqrt{(\Delta x)^2 + (\Delta y)^2}\). Applying the Pythagorean theorem at an infinitesimal scale and summing via integration gives the arc length formula.

・Length of Parametrically Defined Curves
When a curve is defined parametrically by \(x = f(t), y = g(t)\) (\(\alpha \leqq t \leqq \beta\)), its length \(L\) is:
\(L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\)

・Length of Curves in the Form \(y = f(x)\)
The length \(L\) of the curve \(y = f(x)\) (\(a \leqq x \leqq b\)) corresponds to setting \(t = x\) in the parametric formula above (since \(\frac{dx}{dx} = 1\)):
\(L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_a^b \sqrt{1 + \{f'(x)\}^2} dx\)

[Common Mistake] Be Careful When Removing the Square Root!

[Common Mistake]
When finding arc length, the expression inside the radical often turns into a perfect square, producing the form \(\sqrt{(A)^2}\).
Do not blindly drop the radical as \(\sqrt{A^2} = A\). You must treat it with absolute values as \(\sqrt{A^2} = |A|\), check whether the quantity inside is positive or negative within the interval of integration, and resolve the sign properly.

Summary: Putting It All Together

Let's review the key takeaways from Mathematics III Integration:

・Integrals of Standard Functions: Accurately memorize the formulas for trigonometric, exponential, logarithmic, and rational functions, always keeping in mind that integration is the reverse process of differentiation.
・Integration by Substitution: Master linear substitutions of the form \(ax + b = t\), fractions where the numerator is the derivative of the denominator, and the \(x = a \sin \theta\) substitution for definite integrals. Don't forget to transform the integration bounds for definite integrals!
・Integration by Parts: Carefully choose which part to differentiate and which to integrate, executing the pattern "leave \(\times\) integrate \(-\) \(\int\)(differentiate \(\times\) integrate)" smoothly.
・Area, Volume, and Arc Length: Keep the geometric intuition clear in mind: area is "top \(-\) bottom", volume is the "accumulation of cross sections \(S(x)\)", and arc length is "summing microscopic Pythagorean hypotenuses."

Your skill in calculus directly reflects the number of problems you practice. It's totally fine to look at the formulas at first! Work through the standard textbook and workbook exercises repeatedly until the procedures become second nature!