2. Calculus — Integration

Integration by substitution

Integration by substitution (the change of variable method) transforms complex integrals into standard forms by leveraging the chain rule of differentiation in reverse. This topic covers this core calculus technique.

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

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Definition and statement

Theorem of Integration by Substitution (Change of Variable)

Let $g: [a, b] \to \mathbb{R}$ be a differentiable function whose derivative $g'$ is continuous on $[a, b]$, and let $f$ be continuous on the range of $g$. Then, $$\int_{a}^{b} f(g(x)) g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$ where $u = g(x)$ and $du = g'(x)\,dx$. For indefinite integrals: $$\int f(g(x)) g'(x)\,dx = F(g(x)) + C$$ where $F'(u) = f(u)$.

📐 Geometric Interpretation

Geometrically, substitution rescales the domain of integration by matching the infinitesimal differential element $dx$ with the local magnification factor given by the slope $g'(x)$, thereby preserving the net accumulated area under the transformed curve.

⚙️ Algebraic & Structural Meaning

Algebraically, it acts as the inverse operation to the chain rule of differentiation. It linearizes or simplifies integrands containing a function and its derivative product factor, reducing non-standard integrals into elementary antiderivatives.

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 $f$ on the range	`Range of $g(x)$ (Connected interval)`	Strictly Mandatory / Necessary	If $f$ is discontinuous within the range of integration, the antiderivative $F$ may not exist or Riemann integrability fails.
Continuity and Differentiability of $g$	`[a, b] (Closed interval)`	Strictly Mandatory	If $g'(x)$ has jump discontinuities or fails to exist at interior points, the differential transformation $du = g'(x)dx$ becomes invalid.
Monotonicity / Bijectivity of $g$ (for definite integrals)	`[a, b]`	Essential for unique inverse mapping	If $g$ is not injective, multiple domains map to the same range value, leading to overlapping accumulation of area and incorrect definite integral evaluation unless broken into monotonic sub-intervals.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Logarithmic Derivative Pattern

$$\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$$

Rule: If the numerator is the exact derivative of the denominator, the answer is immediately the natural logarithm of the denominator.

Power-Chain Formula

$$\int [f(x)]^n f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)$$

Rule: When a composite function raised to a power is multiplied by its base's derivative, ignore the derivative and increment the power by 1.

Exponential Composite Rule

$$\int e^{f(x)} f'(x)\,dx = e^{f(x)} + C$$

Rule: Whenever an exponential function has an exponent whose derivative multiplies the term, the integral is simply the exponential term itself.

Inverse Trigonometric Differential Forms

$$\int \frac{f'(x)}{\sqrt{1 - [f(x)]^2}}\,dx = \sin^{-1}(f(x)) + C$$

Rule: Identify standard derivatives inside standard inverse trigonometric denominators to instantly write standard inverse trig outputs.

Linear Scaling Substitution

$$ \int f(ax + b)\,dx = \frac{1}{a} F(ax + b) + C $$

Rule: For linear inner functions, perform standard integration with respect to the linear expression and divide by the coefficient of $x$.

Rapid Elimination via Domain & Limits

$$ \int_{a}^{b} f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du $$

Rule: For definite integrals with substitution, always transform limits $u_1 = g(a)$ and $u_2 = g(b)$ instantly. Options failing to change limits can be eliminated at a glance.

Application: Saves up to 45 seconds per question in competitive math papers.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting to Change Limits in Definite Integrals

Common Pitfall: Candidates substitute $u = g(x)$, integrate with respect to $u$, and then plug in the original limits of $x$ without back-substituting or changing limits.

Counterexample: $$\int_{0}^{\pi/2} \sin x \cos x \, dx$$

Why it fails: Plugging $x$-limits into a $u$-expression yields completely erroneous values because $u$ spans a different interval than $x$.

🛡️ Exam Defense: Either change the lower and upper limits immediately upon substitution ($u = g(a)$ and $u = g(b)$) or completely revert to $x$ before applying limits.

⚠️ COMMON ERROR #02

Non-Injective Substitution Pitfall

Common Pitfall: Using a non-monotonic substitution (such as $u = \sin x$ over an interval spanning multiple quadrants) in a definite integral without splitting the domain.

Counterexample: $$\int_{0}^{\pi} \cos x \, dx$$

Why it fails: If $u = \sin x$, $du = \cos x \, dx$, but $\sin x$ is not one-to-one on $[0, \pi]$, causing sign cancellation errors and division-by-zero anomalies in inverse mappings.

🛡️ Exam Defense: Ensure the substitution function $g(x)$ is strictly monotonic (one-to-one) on the interval of integration.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Algebraic Substitution	Continuous derivative $g'(x)$	$$\int f(g(x))g'(x)dx = \int f(u)du$$	Rescaling of differential length element	Presence of function and its derivative product
Logarithmic Form	$f(x) \neq 0$ on interval	$$\int \frac{f'(x)}{f(x)}dx = \ln\|f(x)\| + C$$	Area under hyperbolic rate growth	Rational function with derivative in numerator
Trigonometric Substitution	Invertible trig map ($x = a\sin\theta$)	$$\int \sqrt{a^2 - x^2}dx$$	Circular or hyperbolic arc parametrization	Radical terms like $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$
Integration by Parts	Products of differentiable functions	$$\int u dv = uv - \int v du$$	Area of rectangles (Product rule reversal)	Product of distinct transcendental and algebraic functions

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the definite integral $$\int_{0}^{1} \frac{x^3}{1 + x^8} \, dx$$.

(A) $\frac{\pi}{8}$(B) $\frac{\pi}{4}$(C) $\frac{\pi}{16}$(D) $\frac{1}{4}\ln 2$

📚 Method A: Formal Textbook Derivation

Step 1: Notice $x^8 = (x^4)^2$ and the numerator contains $x^3$, which is proportional to the derivative of $x^4$. Step 2: Put $u = x^4$, so $du = 4x^3 dx \implies x^3 dx = \frac{1}{4} du$. Step 3: Change limits: when $x=0$, $u=0$; when $x=1$, $u=1$. Step 4: The integral becomes $\frac{1}{4} \int_{0}^{1} \frac{1}{1 + u^2} du = \frac{1}{4} [\tan^{-1} u]_{0}^{1} = \frac{1}{4} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{16}$.

⚡ Method B: 25-Second Exam Shortcut

Spot $x^3$ as $\frac{1}{4}\frac{d}{dx}(x^4)$. Immediately write $\frac{1}{4}[\tan^{-1}(x^4)]_0^1 = \frac{1}{4}\frac{\pi}{4} = \frac{\pi}{16}$.

💡 Tip: Always check if the power in the denominator is the square of a variable whose derivative sits in the numerator.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Find the value of $$\int \frac{\dx}{x(\ln x)^3}$$.

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

📚 Method A: Formal Textbook Derivation

Step 1: Recognize that $\frac{1}{x} dx$ is the differential of $\ln x$. Step 2: Let $u = \ln x$, then $du = \frac{1}{x} dx$. Step 3: Integral transforms to $\int u^{-3} du$. Step 4: Integrate using power rule: $\frac{u^{-2}}{-2} + C = -\frac{1}{2(\ln x)^2} + C$.

⚡ Method B: 25-Second Exam Shortcut

Rewrite as $\int (\ln x)^{-3} d(\ln x)$ and apply reverse power rule instantly.

💡 Tip: Logarithmic denominators with powers are frequently asked in UP PGT to test substitution fluency.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Evaluate $$\int x e^{x^2} \, dx$$.

A $e^{x^2} + C$

B $\frac{1}{2} e^{x^2} + C$

C $2 e^{x^2} + C$

D $x e^{x^2} + C$

Derivation & Analysis:

Let $u = x^2$, then $du = 2x dx \implies x dx = \frac{1}{2} du$. Integral becomes $\frac{1}{2} \int e^u du = \frac{1}{2} e^{x^2} + C$.

MCQ 2: Evaluate $$\int \frac{\sec^2(\sqrt{x})}{\sqrt{x}} \, dx$$.

A $\tan(\sqrt{x}) + C$

B $2 \tan(\sqrt{x}) + C$

C $\frac{1}{2} \tan(\sqrt{x}) + C$

D $\sec(\sqrt{x}) + C$

Derivation & Analysis:

Let $u = \sqrt{x} = x^{1/2}$, then $du = \frac{1}{2\sqrt{x}} dx \implies \frac{dx}{\sqrt{x}} = 2 du$. Integral becomes $2 \int \sec^2(u) du = 2 \tan(\sqrt{x}) + C$.

MCQ 3: The value of $$\int_{0}^{\pi/2} \sin^5 x \cos x \, dx$$ is:

A $\frac{1}{6}$

B $\frac{1}{3}$

C $\frac{1}{5}$

D $\frac{1}{2}$

Derivation & Analysis:

Let $u = \sin x$, $du = \cos x dx$. Limits: $x=0 \implies u=0$, $x=\pi/2 \implies u=1$. Integral: $\int_{0}^{1} u^5 du = \left[\frac{u^6}{6}\right]_{0}^{1} = \frac{1}{6}$.

MCQ 4: Evaluate $$\int \frac{1}{x \ln x} \, dx$$.

A $\frac{1}{(\ln x)^2}$

B $\ln|x| + C$

C $\ln|\ln x| + C$

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

Derivation & Analysis:

Let $u = \ln x$, $du = \frac{1}{x} dx$. Integral becomes $\int \frac{1}{u} du = \ln|u| + C = \ln|\ln x| + C$.

MCQ 5: Evaluate $$\int \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \, dx$$.

A $\ln|e^x + e^{-x}| + C$

B $\ln|e^x - e^{-x}| + C$

C $e^x + e^{-x} + C$

D $-\ln|e^x + e^{-x}| + C$

Derivation & Analysis:

Let $u = e^x + e^{-x}$, then $du = (e^x - e^{-x}) dx$. The integral is of the form $\int \frac{du}{u} = \ln|u| + C = \ln|e^x + e^{-x}| + C$.

MCQ 6: Evaluate $$\int \frac{dx}{1 + e^x}$$.

A $x - \ln(1 + e^x) + C$

B $\ln(1 + e^x) + C$

C $-\ln(1 + e^x) + C$

D $\ln(e^x - 1) + C$

Derivation & Analysis:

Rewrite as $\int \frac{e^{-x} dx}{e^{-x} + 1}$. Let $u = e^{-x} + 1$, then $du = -e^{-x} dx$. Integral becomes $-\int \frac{du}{u} = -\ln|e^{-x} + 1| + C = x - \ln(1 + e^x) + C$.

MCQ 7: Evaluate $$\int \frac{\sin(\tan^{-1} x)}{1 + x^2} \, dx$$.

A $\cos(\tan^{-1} x) + C$

B $-\cos(\tan^{-1} x) + C$

C $\sin(\tan^{-1} x) + C$

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

Derivation & Analysis:

Let $u = \tan^{-1} x$, then $du = \frac{dx}{1 + x^2}$. Integral becomes $\int \sin u \, du = -\cos u + C = -\cos(\tan^{-1} x) + C$.

MCQ 8: Assertion (A): $$\int \frac{2x}{(1+x^2)^2} dx = -\frac{1}{1+x^2} + C$$. Reason (R): $\int [f(x)]^n f'(x) dx = \frac{[f(x)]^{n+1}}{n+1} + C$.

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:

Let $u = 1+x^2$, $du = 2x dx$. Integral is $\int u^{-2} du = -u^{-1} + C = -\frac{1}{1+x^2} + C$. Both assertion and reason are true, and reason directly justifies the substitution formula used.

MCQ 9: Evaluate $$\int \frac{\sin x}{\sqrt{1 + \cos x}} \, dx$$.

A $-2\sqrt{1+\cos x} + C$

B $2\sqrt{1+\cos x} + C$

C $-\sqrt{1+\cos x} + C$

D $\frac{1}{2}\sqrt{1+\cos x} + C$

Derivation & Analysis:

Let $u = 1 + \cos x$, then $du = -\sin x dx$. Integral becomes $-\int u^{-1/2} du = -2u^{1/2} + C = -2\sqrt{1+\cos x} + C$.

Q10

MCQ 10: Evaluate $$\int \frac{dx}{\sqrt{x} + x}$$.

A $2 \ln(1 + \sqrt{x}) + C$

B $\ln(1 + \sqrt{x}) + C$

C $2 \ln|x| + C$

D $\frac{1}{2}\ln(1+\sqrt{x}) + C$

Derivation & Analysis:

Factor denominator: $\sqrt{x}(1 + \sqrt{x})$. Let $u = \sqrt{x}$, then $du = \frac{dx}{2\sqrt{x}} \implies \frac{dx}{\sqrt{x}} = 2 du$. Integral becomes $2 \int \frac{du}{1+u} = 2 \ln|1+u| + C = 2 \ln(1+\sqrt{x}) + C$.

Common questions

Look for the inner function of a composite expression, the argument of a transcendental function (trig, exponential, logarithmic), or a denominator whose derivative (or a constant multiple of it) appears elsewhere in the integrand.

While back-substituting into the antiderivative and using original $x$-limits yields the correct answer, changing limits immediately prevents algebraic errors, eliminates redundant back-substitution steps, and saves crucial time in timed examinations like UP PGT.

Memorize standard derivative-integral pairs (like $\frac{f'}{f}$, $f'e^f$, $[f]^n f'$), perform mental substitutions for linear scalings ($ax+b$), and leverage option elimination by checking domain boundaries and derivatives of options.

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