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2. Calculus — Integration

Methods of integration

Methods of integration form the foundational backbone of integral calculus, transforming complex antiderivative problems through algebraic substitution, integration by parts, and partial fraction decomposition. This topic covers these techniques.

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Definition and statement

Fundamental Theorem of Substitution & Integration by Parts
Let $f$ be integrable on $[a, b]$ and let $g: [alpha, \beta] \to [a, b]$ be continuously differentiable. Then $$\int_{g(\alpha)}^{g(\beta)} f(x)\,dx = \int_{\alpha}^{\beta} f(g(t)) g'(t)\,dt$$. Furthermore, if $u(x)$ and $v(x)$ are differentiable functions on an interval, then the integration by parts formula is given by $$\int u \, dv = uv - \int v \, du$$.

📐 Geometric Interpretation

Substitution represents a scaling and re-parametrization of the differential area element $dx$ under a coordinate transformation $x = g(t)$, preserving the total accumulated area under the curve. Integration by parts reflects the product rule for differentiation, geometrically corresponding to the transformation of rectangular area partitions into complementary strip components.

⚙️ Algebraic & Structural Meaning

These methods serve as structural inverses to the chain rule and product rule in differential calculus, allowing the linearization or simplification of transcendental, rational, and irrational algebraic integrands into standard antiderivative forms.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Continuity of the Integrand ($f \in C[a, b]$) [a, b] (Closed interval) Strictly Mandatory / Necessary If $f$ has essential discontinuities or unbounded jumps, Riemann integrability fails unless improper integral limits are invoked.
Differentiability of the Substitution Function ($g \in C^1[\alpha, \beta]$) [\alpha, \beta] (Closed interval) Strictly Mandatory Non-smooth or non-monotonic substitution functions $g(t)$ result in non-invertible transformations, yielding multi-valued domains or incorrect measure scaling.
Differentiability of Factors in Integration by Parts (a, b) (Open interval) Essential for conclusion If $u$ or $v$ fails to be differentiable, the product rule expansion $\frac{d}{dx}(uv) = u'v + uv'$ is invalid, breaking the integral identity.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Linear Combination Exponential Integrals (Euler's Form)

$$ \int e^{ax} \cos(bx)\,dx = \frac{e^{ax}}{a^2 + b^2} [a \cos(bx) + b \sin(bx)] + C $$

Rule: Directly write the result without performing integration by parts twice. The denominator is always $a^2 + b^2$, and the numerator cycles $a$ with the trigonometric function and $b$ with its co-function.

02

King's Property for Definite Integrals

$$ \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx $$

Rule: When the sum of limits $a+b$ simplifies the argument $f(a+b-x)$ to cancel terms or combine with $f(x)$ into a constant (often $1$ or $\pi/2$), add the original and transformed integrals to solve instantly.

03

Reduction Formulae for Powers of Sine and Cosine (Wallis's Formula)

$$ \int_{0}^{\pi/2} \sin^n x \, dx = \int_{0}^{\pi/2} \cos^n x \, dx = \frac{(n-1)(n-3)\dots}{n(n-2)\dots} \cdot k $$ where $k = \pi/2$ if $n$ is even, and $1$ if $n$ is odd.

Rule: Instantly evaluate definite integrals of powers of $\sin x$ or $\cos x$ from $0$ to $\pi/2$ without manual antidifferentiation.

04

Rational Integrals with Quadratic Denominators ($1/(ax^2+bx+c)$)

$$ \int \frac{dx}{ax^2 + bx + c} = \frac{2}{\sqrt{4ac - b^2}} \tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac - b^2}}\right) + C \quad (4ac > b^2) $$

Rule: Skip completing the square by plugging coefficients directly into the standard inverse tangent or logarithmic derivative formulae.

05

Integrals of the Form $\int e^x [f(x) + f'(x)] \, dx$

$$ \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C $$

Rule: Spot the function and its exact derivative multiplied by $e^x$; write the product $e^x f(x)$ immediately.

06

Leibniz Rule for Differentiation Under Integral Sign

$$ \frac{d}{dx} \left[ \int_{\psi(x)}^{\phi(x)} f(t)\,dt \right] = f(\phi(x)) \cdot \phi'(x) - f(\psi(x)) \cdot \psi'(x) $$

Rule: Differentiate parametric definite integrals instantly by substituting upper/lower limits into the integrand and multiplying by their respective derivatives.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Domain Restrictions in Logarithmic Integrals

Common Pitfall: Writing $\int \frac{1}{x} \, dx = \ln x + C$ without absolute value bars, or failing to check the domain of the integrand.

Counterexample: $$\int_{-2}^{-1} \frac{1}{x} \, dx = [\ln |x|]_{-2}^{-1} = \ln 1 - \ln 2 = -\ln 2$$

Why it fails: Omitting absolute value bars leads to undefined real logarithms for negative integration domains ($x < 0$).

🛡️ Exam Defense: Always write $\ln |f(x)| + C$ for $\int \frac{f'(x)}{f(x)} \, dx$, and check interval boundaries carefully.
⚠️ COMMON ERROR #02

Blind Application of King's Property on Non-Symmetric Intervals

Common Pitfall: Applying $\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx$ when the lower limit is not zero or when the interval is not symmetric, leading to algebraic tangles.

Counterexample: $$\int_{1}^{3} x^2 \, dx \neq \int_{1}^{3} (4-x)^2 \, dx \text{ without proper offset tracking}$$

Why it fails: The substitution $t = a+b-x$ shifts the differential $dt = -dx$ and flips limits; ignoring this sign change creates sign errors.

🛡️ Exam Defense: Verify that the integrand exhibits structural symmetry around the midpoint $(a+b)/2$ before attempting substitution shortcuts.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Integration by Substitution$g \in C^1$, $f$ continuous$\int f(g(t))g'(t)dt = \int f(x)dx$Area preservation under coordinate stretchingPresence of both a function and its derivative
Integration by Parts$u, v$ differentiable on interval$\int u dv = uv - \int v du$Transformation of rectangular area partitionsProduct of distinct function families (Algebraic $\times$ Trig/Log)
Partial FractionsRational function $P(x)/Q(x)$, $\text{deg}(P) < \text{deg}(Q)$$\sum \frac{A_i}{factor_i}$Decomposition of complex area into elementary logarithmic/rational stripsFactorizable polynomial denominator
King's Property$f$ continuous on $[a, b]$$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$Reflection symmetry of area about midpointDefinite integrals with trigonometric or logarithmic terms in limits $[0, \pi/2]$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the definite integral $$ \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx $$.

(A) $\frac{\pi}{4}$(B) $\frac{\pi}{2}$(C) $\frac{\pi}{3}$(D) $\frac{\pi}{6}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Let $I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$. Step 2: Apply King's property $\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx$ to get $I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx$. Step 3: Add both equations: $2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}$. Step 4: Therefore, $I = \frac{\pi}{4}$.
⚡ Method B: 25-Second Exam Shortcut
Recognize the standard symmetry form $\frac{f(x)}{f(x) + f(a+b-x)}$ on interval $[a, b]$. The answer is always $\frac{b-a}{2} = \frac{\pi/2 - 0}{2} = \frac{\pi}{4}$.
💡 Tip: Whenever the denominator is the sum of complementary trigonometric functions, the answer is invariably $\pi/4$ if limits are $0$ to $\pi/2$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Evaluate the indefinite integral $$ \int e^{x} \left(\frac{1 + x \ln x}{x}\right) \, dx $$.

(A) $e^x \ln x + C$(B) $e^x \ln x + \frac{e^x}{x} + C$(C) $e^x (x \ln x) + C$(D) $\frac{e^x \ln x}{x} + C$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Rewrite the integrand as $\int e^x \left(\frac{1}{x} + \ln x\right) \, dx$. Step 2: Let $f(x) = \ln x$. Then $f'(x) = \frac{1}{x}$. Step 3: Apply the standard formula $\int e^x [f(x) + f'(x)]dx = e^x f(x) + C$. Step 4: Result is $e^x \ln x + C$.
⚡ Method B: 25-Second Exam Shortcut
Separate terms into $\frac{1}{x}$ and $\ln x$; observe that derivative of $\ln x$ is $\frac{1}{x}$, yielding immediate solution $e^x \ln x + C$.
💡 Tip: Look for logarithmic and reciprocal pairings multiplied by $e^x$ in PGT exams.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: Evaluate the integral $$\int \frac{dx}{x(x^5 + 1)}$$
A $\frac{1}{5} \ln \left| \frac{x^5}{x^5 + 1} \right| + C$
B $\ln |x^5 + 1| + C$
C $\frac{1}{5} \ln |x^5 + 1| + C$
D $-\frac{1}{5} \ln |x^5 + 1| + C$
Derivation & Analysis:

Multiply numerator and denominator by $x^4$ to get $$\int \frac{x^4 dx}{x^5(x^5 + 1)}$$. Put $x^5 = t$, so $5x^4 dx = dt$. The integral becomes $$\frac{1}{5} \int \frac{dt}{t(t+1)} = \frac{1}{5} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt = \frac{1}{5} \ln \left| \frac{t}{t+1} \right| + C = \frac{1}{5} \ln \left| \frac{x^5}{x^5 + 1} \right| + C$$.

Q2
MCQ 2: The value of $$\int_{-1}^{1} \frac{sin^5 x \cos^4 x}{1 + x^2} \, dx$$ is:
A $\frac{\pi}{2}$
B $1$
C $0$
D $\frac{1}{2}$
Derivation & Analysis:

Let $f(x) = \frac{\sin^5 x \cos^4 x}{1 + x^2}$. Check parity: $f(-x) = \frac{\sin^5(-x) \cos^4(-x)}{1 + (-x)^2} = \frac{-\sin^5 x \cos^4 x}{1 + x^2} = -f(x)$. Since the integrand is an odd function and the interval $[-1, 1]$ is symmetric about the origin, the definite integral evaluates identically to $0$.

Q3
MCQ 3: Evaluate $$\int \frac{dx}{\sin(x - a) \sin(x - b)}$$
A $\csc(a - b) \ln \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + C$
B $\sin(a - b) \ln \left| \frac{\sin(x - b)}{\sin(x - a)} \right| + C$
C $\ln |\sin(x - a) \sin(x - b)| + C$
D $\tan(a - b) \ln |\sin(x - a)| + C$
Derivation & Analysis:

Multiply and divide by $\sin(a - b)$: $$\frac{1}{\sin(a-b)} \int \frac{\sin((x-b) - (x-a))}{\sin(x-a)\sin(x-b)} dx$$. Expanding numerator gives $$\frac{1}{\sin(a-b)} \int [\cot(x-b) - \cot(x-a)] dx = \frac{1}{\sin(a-b)} \ln \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C$$ (adjusting sign gives option A).

Q4
MCQ 4: What is the value of $$\int_{0}^{\pi/2} \ln(\tan x) \, dx$$?
A $\pi$
B $\frac{\pi}{2}$
C $1$
D $0$
Derivation & Analysis:

Let $I = \int_{0}^{\pi/2} \ln(\tan x) \, dx$. By King's property, $I = \int_{0}^{\pi/2} \ln(\tan(\frac{\pi}{2} - x)) \, dx = \int_{0}^{\pi/2} \ln(\cot x) \, dx$. Adding both: $2I = \int_{0}^{\pi/2} [\ln(\tan x) + \ln(\cot x)] dx = \int_{0}^{\pi/2} \ln(1) dx = 0$. Thus $I = 0$.

Q5
MCQ 5: Evaluate $$\int \frac{x^2 + 1}{(x^2 - 1)^2} \, dx$$
A $\frac{-x}{x^2 - 1} + C$
B $\frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C$
C $\frac{-2x}{x^2 - 1} + C$
D $\ln |x^2 - 1| + C$
Derivation & Analysis:

Divide numerator and denominator by $x^2$: $$\int \frac{1 + 1/x^2}{(x - 1/x)^2} dx$$. Let $x - 1/x = t$, so $(1 + 1/x^2)dx = dt$. The integral becomes $$\int \frac{dt}{t^2} = -\frac{1}{t} + C = -\frac{1}{x - 1/x} + C = \frac{-x}{x^2 - 1} + C$$.

Q6
MCQ 6: The integral $$\int \frac{dx}{\sqrt{x(1 - x)}}$$ equals:
A $2 \sin^{-1}(\sqrt{x}) + C$
B $\sin^{-1}(2x - 1) + C$
C $\ln |\sqrt{x} + \sqrt{1-x}| + C$
D $2 \ln |\sqrt{x} + \sqrt{1-x}| + C$
Derivation & Analysis:

Rewrite the denominator as $\sqrt{x - x^2} = \sqrt{1/4 - (x - 1/2)^2}$. Alternatively, substitute $x = \sin^2 \theta$, so $dx = 2 \sin \theta \cos \theta \, d\theta$. The integral transforms to $$\int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \cos \theta} = \int 2 \, d\theta = 2\theta + C = 2 \sin^{-1}(\sqrt{x}) + C$$.

Q7
MCQ 7: If $I_n = \int \tan^n x \, dx$, then $I_n + I_{n-2}$ is equal to:
A $\frac{\tan^{n-1} x}{n-1} + C$
B $\frac{\tan^{n-1} x}{n} + C$
C $\tan^{n-1} x + C$
D $\frac{-\tan^{n-1} x}{n-1} + C$
Derivation & Analysis:

We have $I_n = \int \tan^{n-2}x (\sec^2 x - 1) dx = \int \tan^{n-2}x \sec^2 x dx - I_{n-2}$. Evaluating the first part by substitution ($\tan x = t$) yields $$\frac{\tan^{n-1}x}{n-1}$$. Thus, $I_n + I_{n-2} = \frac{\tan^{n-1}x}{n-1} + C$.

Q8
MCQ 8: Assertion (A): $$\int_{-a}^{a} f(x)dx = 2 \int_{0}^{a} f(x)dx$$ if $f$ is even. Reason (R): Definite integrals accumulate signed area across symmetric boundaries.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

For an even function $f(-x) = f(x)$, the geometric area from $-a$ to $0$ equals the area from $0$ to $a$ due to y-axis symmetry, validating both the property and its geometric rationale.

Q9
MCQ 9: Evaluate $$\int \frac{e^{x}(1 + \sin x)}{1 + \cos x} \, dx$$
A $e^x \tan\left(\frac{x}{2}\right) + C$
B $e^x \cot\left(\frac{x}{2}\right) + C$
C $e^x \sec^2\left(\frac{x}{2}\right) + C$
D $e^x \ln(1 + \cos x) + C$
Derivation & Analysis:

Use half-angle identities: $1 + \sin x = (\sin(x/2) + \cos(x/2))^2$ and $1 + \cos x = 2 \cos^2(x/2)$. The integrand becomes $$\frac{1}{2} e^x \left[ \tan^2(x/2) + 1 + 2\tan(x/2) \right]$$ which simplifies to $e^x \left[ \tan(x/2) + \frac{1}{2}\sec^2(x/2) \right]$. Letting $f(x) = \tan(x/2)$, its derivative is $\frac{1}{2}\sec^2(x/2)$, yielding $e^x \tan(x/2) + C$.

Q10
MCQ 10: The value of $$\int_{0}^{\pi} x \sin^3 x \, dx$$ is:
A $\frac{2\pi}{3}$
B $\frac{\pi}{3}$
C $\frac{4\pi}{3}$
D $\pi$
Derivation & Analysis:

Let $I = \int_{0}^{\pi} x \sin^3 x \, dx$. Using property $\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx$, we get $I = \int_{0}^{\pi} (\pi - x)\sin^3(\pi - x)dx = \int_{0}^{\pi} (\pi - x)\sin^3 x \, dx$. Adding both equations: $2I = \pi \int_{0}^{\pi} \sin^3 x \, dx = \pi \cdot 2 \int_{0}^{\pi/2} \sin^3 x \, dx = 2\pi \left(\frac{2}{3}\right) = \frac{4\pi}{3}$. Thus $I = \frac{2\pi}{3}$.

Common questions

Look for compositional structures: if an integrand contains a function and its derivative multiplied together ($f(g(x))g'(x)$), always use substitution. If the integrand is a product of two distinct algebraic families (e.g., polynomial $\times$ trigonometric, or logarithmic $\times$ algebraic), apply integration by parts using LIATE (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) priority.

Failure to account for domain restrictions (such as missing absolute value bars in logarithmic integrals) and incorrect handling of limits when applying definite integral properties like King's theorem or odd/even parity.

Memorize standard symmetry results (such as $\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx = \frac{\pi}{4}$) and master King's Property ($a+b-x$) substitution. Recognizing these structural templates allows you to bypass lengthy algebraic derivations entirely.

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