2. Calculus — Integration

Integration by parts

Integration by parts is the foundational integral calculus counterpart to the product rule for differentiation, converting the integral of a product into alternative forms.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Integration by Parts Formula

If $u = f(x)$ and $v = g(x)$ are two differentiable functions of $x$, then the integral of their product is given by $$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x)\int g(x)\,dx\right]dx$$, or in compact notation, $$\int u \, dv = u v - \int v \, du$$.

📐 Geometric Interpretation

Geometrically, integration by parts relates the area under a product curve to complementary rectangular sub-regions and the area under the transformed derivative curve, underpinning transformations in coordinate geometry.

⚙️ Algebraic & Structural Meaning

It stems directly from the differential product rule $d(uv) = u\,dv + v\,du$, serving as an indispensable tool for reducing the algebraic complexity of transcendental and polynomial-algebraic mixed integrands.

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 Functions	`[a, b] (Closed interval)`	Strictly Mandatory / Necessary	If $f(x)$ or $g(x)$ is discontinuous on $[a, b]$, the Riemann integral may not exist or the antiderivative cannot be evaluated continuously.
Differentiability of the first factor	`(a, b) (Open interval)`	Strictly Mandatory	The derivative $f'(x)$ must exist on the interval, otherwise the secondary integral containing $f'(x)$ cannot be formed.
Integrability of the second factor	`[a, b] (Closed interval)`	Essential for conclusion	If $\int g(x)\,dx$ is undefined or diverges, the integration by parts reduction collapses.

Useful results and shortcuts

Use these identities and results when their conditions apply:

The LIATE Priority Rule for Choosing $u$

$$ \text{LIATE: } \text{Logarithmic} \to \text{Inverse Trig} \to \text{Algebraic} \to \text{Trigonometric} \to \text{Exponential} $$

Rule: Always assign $u$ to the function appearing earlier in the LIATE acronym to ensure algebraic simplification upon differentiation.

Application: Directly applied in UP PGT/TGT to integrate $\int x^n \ln x \, dx$ or $\int x^2 e^x \, dx$.

Bernoulli's Tabular Integration (DI Method)

$$ \int P(x)e^{ax}\,dx = P(x)\frac{e^{ax}}{a} - P'(x)\frac{e^{ax}}{a^2} + P''(x)\frac{e^{ax}}{a^3} - \cdots $$

Rule: For polynomials multiplied by exponentials or sines/cosines, differentiate the polynomial successively to zero while integrating the other part in parallel with alternating signs.

Application: Crucial time-saver for degree-3 or degree-4 polynomial integrals frequently asked in DSSSB PGT and UP PGT.

Exponential-Trigonometric Cyclic Integral

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

Rule: Recognize the cyclic loop after applying parts twice and solve algebraically without recalculating.

Application: Standard recurring question type in KVS PGT and UP TGT mathematics.

Reduction Formula for Powers of Sines and Cosines

$$ \int \sin^n x \, dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x \, dx $$

Rule: Use recursive reduction steps to drop high integer powers down to standard forms instantly.

Application: Frequently featured in LT Grade and UP PGT definitive calculus problems.

Integral of $e^x[f(x) + f'(x)]$

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

Rule: Spot the function and its derivative combined with $e^x$ to write the answer instantly.

Rapid Elimination via Logarithmic Arguments

$$ \int \ln(x^2 + a^2)\,dx = x\ln(x^2+a^2) - 2x + 2a\tan^{-1}\left(\frac{x}{a}\right) + C $$

Rule: Memorize standard inverse and logarithmic integral expansion structures to eliminate incorrect sign options immediately.

Application: Used to eliminate 2 distractor options in multi-term logarithmic integrals.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting the Integration Constant in Intermediate Steps

Common Pitfall: Omitting $+ C$ during intermediate integration of $dv$ can lead to compounding algebraic errors in definite or cyclic integrals.

Counterexample: $$\int x \cdot 0 \, dx \neq \text{incomplete forms}$$

Why it fails: In cyclic integrals like $\int e^x \cos x \, dx$, missing or mishandling constants leads to self-cancelling identity traps ($0 = 0$).

🛡️ Exam Defense: Combine or properly account for constants only at the final stage of indefinite integration.

⚠️ COMMON ERROR #02

Wrong Choice of $u$ and $dv$

Common Pitfall: Choosing an algebraic function as $dv$ when it increases in complexity upon integration (e.g., trying to integrate $\ln x$ by setting $u = x$).

Counterexample: $$\int \ln x \, dx \text{ with } u = x, dv = \frac{1}{x}dx \text{ leads to non-termination}$$

Why it fails: The resulting integral becomes more complex rather than simpler.

🛡️ Exam Defense: Always follow the LIATE hierarchy strictly.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Integration by Parts	Differentiable $u$, Integrable $v$	$uv - \int v\,du$	Area balance of product regions	Product of dissimilar function types
Bernoulli's Tabular Method	Polynomial $\times$ Exponential/Trig	Alternating sum of derivatives/integrals	Repeated application shorthand	Polynomial degree $n \ge 2$ with $e^{ax}$
Cyclic Integration by Parts	Exponential with Sine/Cosine	$I = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2}$	Phase-shifted harmonic recurrence	$e^{ax}\sin bx$ or $e^{ax}\cos bx$
Standard Derivative Product Rule	Differentiable $uv$	$d(uv) = u\,dv + v\,du$	Infinitesimal rectangle area expansion	Differential verification

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the integral $I = \int x e^{2x} \, dx$.

(A) \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C(B) x e^{2x} - e^{2x} + C(C) \frac{1}{4}x e^{2x} - \frac{1}{2}e^{2x} + C(D) \frac{1}{2}x e^{2x} + \frac{1}{4}e^{2x} + C

📚 Method A: Formal Textbook Derivation

Let $u = x$ and $dv = e^{2x}dx$. Then $du = dx$ and $v = \frac{1}{2}e^{2x}$. Applying parts: $\int x e^{2x}dx = x \left(\frac{1}{2}e^{2x}\right) - \int \frac{1}{2}e^{2x}dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$.

⚡ Method B: 25-Second Exam Shortcut

Use Bernoulli's table: Differentiate $x \to 1 \to 0$; Integrate $e^{2x} \to \frac{1}{2}e^{2x} \to \frac{1}{4}e^{2x}$. Combine with signs $(+) (-)$: $x(\frac{1}{2}e^{2x}) - 1(\frac{1}{4}e^{2x}) + C$.

💡 Tip: Check coefficients of exponential denominators carefully to avoid sign traps.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Evaluate $I = \int \ln x \, dx$.

(A) x \ln x + x + C(B) x \ln x - x + C(C) \frac{\ln x}{x} + C(D) \ln x - x + C

📚 Method A: Formal Textbook Derivation

Rewrite as $\int (\ln x) \cdot 1 \, dx$. Let $u = \ln x, dv = dx$. Then $du = \frac{1}{x}dx, v = x$. Formula gives $x \ln x - \int x \cdot \frac{1}{x}dx = x \ln x - \int 1 \, dx = x \ln x - x + C$.

⚡ Method B: 25-Second Exam Shortcut

Direct standard result memorization for logarithmic integrals.

💡 Tip: Frequently asked as a direct formula in UP TGT exams.

Practice questions

Choose an answer to check it and see the solution:

What is the value of $\int x \sin x \, dx$?

A $-x \cos x + \sin x + C$

B $x \cos x - \sin x + C$

C $-x \sin x + \cos x + C$

D $x \sin x + \cos x + C$

Derivation & Analysis:

Set $u = x, dv = \sin x dx$. Then $du = dx, v = -\cos x$. Integral = $-x\cos x - \int(-\cos x)dx = -x\cos x + \sin x + C$.

Evaluate $\int x^2 e^x \, dx$.

A $(x^2 - 2x + 2)e^x + C$

B $(x^2 + 2x + 2)e^x + C$

C $(x^2 - 2x - 2)e^x + C$

D None of these

Derivation & Analysis:

Using tabular integration: Polynomial $x^2 \to 2x \to 2 \to 0$. Exponential $e^x \to e^x \to e^x \to e^x$. Signs: $+ - +$. Result: $x^2(e^x) - 2x(e^x) + 2(e^x) + C = (x^2 - 2x + 2)e^x + C$.

The integral $\int e^x (\tan x + \sec^2 x) \, dx$ equals:

A $e^x \tan x + C$

B $e^x \sec x + C$

C $e^x \ln(\sec x) + C$

D $e^x \cot x + C$

Derivation & Analysis:

Here $f(x) = \tan x$ and $f'(x) = \sec^2 x$. By the standard property $\int e^x [f(x)+f'(x)]dx = e^x f(x) + C$, the answer is $e^x \tan x + C$.

Evaluate $\int \tan^{-1} x \, dx$.

A $x \tan^{-1} x + \frac{1}{2}\ln(1+x^2) + C$

B $x \tan^{-1} x - \frac{1}{2}\ln(1+x^2) + C$

C $\tan^{-1} x - x + C$

D $\frac{x}{1+x^2} + C$

Derivation & Analysis:

Take $u = \tan^{-1}x, dv = dx$. Then $du = \frac{1}{1+x^2}dx, v = x$. Result is $x\tan^{-1}x - \int \frac{x}{1+x^2}dx = x\tan^{-1}x - \frac{1}{2}\ln(1+x^2) + C$.

Evaluate $\int e^{2x} \cos(3x) \, dx$.

A $\frac{e^{2x}}{13}(2\cos 3x + 3\sin 3x) + C$

B $\frac{e^{2x}}{13}(2\cos 3x - 3\sin 3x) + C$

C $\frac{e^{2x}}{13}(3\cos 3x + 2\sin 3x) + C$

D None of these

Derivation & Analysis:

Using formula $\int e^{ax}\cos(bx)dx = \frac{e^{ax}}{a^2+b^2}[a\cos(bx)+b\sin(bx)]+C$ with $a=2, b=3$: $\frac{e^{2x}}{4+9}[2\cos 3x + 3\sin 3x] + C$.

What is $\int \sec^3 x \, dx$?

A $\frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$

B $\sec x \tan x + \ln|\sec x + \tan x| + C$

C $\frac{1}{2}\sec^2 x + C$

D None

Derivation & Analysis:

Writing as $\int \sec x \cdot \sec^2 x dx$, take $u = \sec x, dv = \sec^2 x dx$. Reduction yields $\frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$.

Evaluate $\int x \sec^2 x \, dx$.

A $x \tan x + \ln|\cos x| + C$

B $x \tan x - \ln|\cos x| + C$

C $x \tan x + \tan x + C$

D $x \sec x + C$

Derivation & Analysis:

Let $u = x, dv = \sec^2 x dx \implies v = \tan x$. Integral is $x\tan x - \int \tan x dx = x\tan x - \ln|\sec x| + C = x\tan x + \ln|\cos x| + C$.

Find $\int (x^2 + 1)\ln x \, dx$.

A $\left(\frac{x^3}{3} + x\right)\ln x - \frac{x^3}{9} - x + C$

B $\left(\frac{x^3}{3} + x\right)\ln x + \frac{x^3}{9} + x + C$

C $\frac{x^3}{3}\ln x + C$

D None of these

Derivation & Analysis:

LIATE dictates $u = \ln x, dv = (x^2+1)dx \implies v = \frac{x^3}{3}+x$. Applying parts gives the expanded algebraic form.

If $I_n = \int \tan^n x \, dx$, then $I_n + I_{n-2}$ equals:

A $\frac{\tan^{n-1}x}{n-1}$

B $\frac{\tan^{n-1}x}{n-1} + C$

C $\tan^{n-1}x$

D 0

Derivation & Analysis:

Standard reduction relation for tangent powers gives $I_n = \frac{\tan^{n-1}x}{n-1} - I_{n-2}$, hence $I_n + I_{n-2} = \frac{\tan^{n-1}x}{n-1}$.

Q10

Evaluate $\int x^3 \ln x \, dx$.

A $\frac{x^4}{4}\ln x - \frac{x^4}{16} + C$

B $\frac{x^4}{4}\ln x + \frac{x^4}{16} + C$

C $x^4 \ln x - x^4 + C$

D $\frac{1}{4}x^3 \ln x + C$

Derivation & Analysis:

Let $u = \ln x, dv = x^3 dx \implies v = \frac{x^4}{4}$. Formula: $\frac{x^4}{4}\ln x - \int \frac{x^4}{4}\cdot \frac{1}{x}dx = \frac{x^4}{4}\ln x - \frac{x^4}{16} + C$.

Common questions

Use substitution when an explicit function-derivative pair ($f(x)$ and $f'(x)$) exists within the integrand. Use integration by parts when the integrand is a product of two completely different elementary functions (such as algebraic and trigonometric or logarithmic).

Sign errors during the second integral step—especially when the derivative of the first term or the integral of the second term introduces negative signs—and forgetting the arbitrary constant $+ C$ in indefinite integrals.

Mastering Bernoulli's Tabular method for polynomial-exponential products and memorizing standard cyclic integral results saves 60 to 90 seconds per question.

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