Antiderivatives(II) | NEB Class 11 Mathematics Notes

Chapter Overview

Namaste students! Welcome to these comprehensive notes on Antiderivatives & Techniques of Integration. In Class 11, this is one of the most scoring yet conceptual chapters in Calculus. Think of an antiderivative as differentiation in reverse — hit 'undo' on a derivative, and you are back at the original function. Given a derivative, our goal is to reconstruct the original equation. This foundation unlocks the keys to finding areas under curves, volumes of irregular solids, and solving differential equations.

What this chapter covers

Concept of the Antiderivative and Indefinite IntegralThe Substitution Method (Forms I, II, and III)Trigonometric Substitutions for Radical FormsIntegration by Parts (The ILATE Rule)Standard Integrals of Trigonometric and Inverse Trigonometric Functions

Learning Objectives

1

Apply the Substitution Method to simplify complex algebraic and trigonometric integrals

2

Use Trigonometric Substitutions to evaluate radical forms involving square roots of sum or difference of squares

3

Master the technique of Integration by Parts for products of two functions

4

Employ the ILATE rule systematically to select the first function in product integration

The Core Idea of Integration Techniques

Key Insight

If an integral is not in a standard form, do not panic. Your primary objective in this chapter is to transform the unfamiliar into the familiar using substitutions, trigonometric simplifications, or integration by parts.

Look for patterns. If you see a function and its derivative living in the same expression, substitution is almost always your best path forward.

Key Definitions

Antiderivative

A function $F(x)$ is called an antiderivative of $f(x)$ if the derivative of $F(x)$ with respect to $x$ equals $f(x)$ for all values in the domain.

$$F'(x) = f(x)$$

Indefinite Integral

The mathematical collection of all possible antiderivatives of a function $f(x)$, containing an arbitrary constant of integration $C$.

$$\int f(x) \, dx = F(x) + C$$

Integrand

The specific function $f(x)$ enclosed within the integration operator that is being integrated.

$$\text{In } \int f(x) \, dx, \text{ the integrand is } f(x)$$

Integration by Substitution

The substitution method simplifies an integral by changing the variable from $x$ to $y$ (or $u$). This is used when the integrand is not in standard form but contains a function and its derivative.

$$\int f(g(x)) g'(x) \, dx = \int f(y) \, dy \quad [y = g(x)]$$

Why it works (Derivation):

1Form I — Linear expression: $f(ax+b)$
Put $y = ax+b$. Differentiating gives $dy = a \, dx$, which means $dx = \frac{1}{a} \, dy$. The integral simplifies to $\frac{1}{a} \int f(y) \, dy$.

2Form II — Power patterns: $x^{n-1} f(x^n)$
Put $y = x^n$. Differentiating gives $n x^{n-1} \, dx = dy$, meaning $x^{n-1} \, dx = \frac{1}{n} \, dy$. The expression simplifies to $\frac{1}{n} \int f(y) \, dy$.

3Form III — Functions raised to powers: $\{f(x)\}^n f'(x)$
Put $y = f(x)$. Differentiating gives $f'(x) \, dx = dy$. This simplifies the integral directly into $\int y^n \, dy$.

Example: Evaluate $\int x \cos(x^2+5) \, dx$. Let $y = x^2+5$. Then $dy = 2x \, dx \implies x \, dx = \frac{dy}{2}$. Substituting gives $\frac{1}{2}\int \cos y \, dy = \frac{1}{2}\sin y + C = \frac{1}{2}\sin(x^2+5) + C$.

When to use: Scan the integrand first. If you spot a function paired with its derivative (even if off by a constant factor), substitution is the correct tool.

Substitution Method vs Integration by Parts

Choosing the correct technique of integration is crucial for solving exam questions efficiently.

Integration by Substitution

$$\int f(g(x)) g'(x) \, dx$$

Used when a function and its derivative both exist in the integrand

Simplifies composite functions like $f(ax+b)$ or $f(x^n)$

Relies on variable transformation (e.g., setting $y = g(x)$)

Changes the differential from $dx$ to $dy$

VS

Integration by Parts

$$\int u \, dv = u v - \int v \, du$$

Used when the integrand is a product of two distinct types of functions

Applicable to forms like $x \sin x$, $x e^x$, or $\ln x$

Relies on the derivative product rule reversed

Selection of the first function '$u$' is determined by ILATE order

Rule of Thumb: Always check for substitution first. If you cannot find a function-derivative pairing, and the integrand is a product of different function families, switch to Integration by Parts.

Advanced Techniques & Special Integration Formulas

The Standard Power Integral

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C

Standard rule for non-negative integers or rational powers, where n ≠ -1.

The Logarithmic Reciprocal Rule

\int \frac{1}{x} \, dx = \ln|x| + C

Applicable for single-degree denominators.

Linear Exponential Rule

\int e^{ax} \, dx = \frac{e^{ax}}{a} + C

Remember to divide by the linear coefficient a of the exponent.

Linear Sine Rule

\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C

Result starts with 'co-' so it is negative.

Secant Squared Integral

\int \sec^2 x \, dx = \tan x + C

Direct result of tangent differentiation.

Inverse Tangent Standard Integral

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

Derived using the substitution x = a tan θ.

Inverse Sine Standard Integral

\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C

Derived using the substitution x = a sin θ. Note that there is no 1/a coefficient outside.

Integration by Parts Formula

\int (u \cdot v) \, dx = u \int v \, dx - \int \left[\frac{du}{dx} \cdot \int v \, dx\right] dx

Derivation shown below. Select function u according to ILATE order.

1Start with the differentiation product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.

2Integrate both sides with respect to $x$: $$\int \frac{d}{dx}(uv) \, dx = \int \left[u \frac{dv}{dx} + v \frac{du}{dx}\right] dx$$

3Simplify the left-hand side: $$uv = \int u \frac{dv}{dx} \, dx + \int v \frac{du}{dx} \, dx$$

4Rearrange the terms to isolate the integral of interest: $$\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$$

5Let $\frac{dv}{dx} = g(x)$, meaning $v = \int g(x) \, dx$. Substituting these values gives the final Integration by Parts formula: $$\int f(x)g(x) \, dx = f(x) \int g(x) \, dx - \int \left[f'(x) \int g(x) \, dx\right] dx$$

Tricks for Choosing Substitutions & ILATE

Trick: Two simple mental models that make integration by parts and trigonometric substitution intuitive.

Steps to Remember:

ILATE Selection Rule: When integrating a product, match functions against the word ILATE from left to right. The function type that appears first is chosen as $u$, and the remaining function is $v$:
• I: Inverse Trigonometric (e.g., $\sin^{-1} x$)
• L: Logarithmic (e.g., $\ln x$)
• A: Algebraic (e.g., $x^2$, $3x$)
• T: Trigonometric (e.g., $\sin x$)
• E: Exponential (e.g., $e^x$)
Trigonometric Substitution Unlock: If you see algebraic radicals with $x^2$ and $a^2$, use these substitutions to simplify:
• For $\sqrt{a^2-x^2} \implies$ substitute $x = a\sin\theta$ or $x = a\cos\theta$
• For $\sqrt{a^2+x^2} \implies$ substitute $x = a\tan\theta$ or $x = a\cot\theta$
• For $\sqrt{x^2-a^2} \implies$ substitute $x = a\sec\theta$ or $x = a\csc\theta$
The Derivative Match tip: Always scan the integrand for function-derivative pairs. If you see $\ln x$, check for a $\frac{1}{x}$ divisor. If you see $x^2$, check for an $x$ variable outside.

Mnemonic: ILATE: Inverse, Log, Algebra, Trig, Exp | Trig Substitution: a²-x² uses sin, a²+x² uses tan, x²-a² uses sec

Common Mistakes to Avoid

Avoiding these three common errors will keep you from losing marks on exam day.

Forgetting the constant of integration (+C): Always include $+C$ on indefinite integrals. It is a standard mathematical requirement representing the infinite family of curves.
Choosing the wrong 'u' in Integration by Parts: Violating ILATE can make your integral significantly more complex or impossible to solve. Always verify the hierarchy before expanding.
Variable mismatch inside substitution: When substituting $y = f(x)$, always transform $dx$ to $dy$ before integrating. Working with variables like $\int y \, dx$ is mathematically invalid.

NEB Exam Booster — Crucial Rules

NEB Exam Warning

The following techniques and special identities are highly tested by NEB examiners.

The Logarithmic Derivative shortcut: Keep this rule in mind: $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$. It instantly evaluates fractional expressions.
Inverse Trig substitutions: If your integrand contains an inverse trigonometric function on its own (like $\int \sin^{-1} x \, dx$), treat it as a product by writing $1 \cdot \sin^{-1} x \, dx$, then use Integration by Parts.
ILATE Hierarchy: Never skip checking the order. Logarithmic and Inverse Trig are almost always selected as $u$ over algebraic and trigonometric functions.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Linear compound expressions inside brackets like (5x+1)³	Apply linear integration substitution (Form I).
Products of two different function types (e.g., x · eˣ)	Use Integration by Parts with ILATE to determine u and v.
Radicals like √(a² - x²)	Apply trigonometric substitution: set x = a sin θ.
Derivative of the denominator is present in the numerator	Integrate directly using ln\|denominator\| + C.

Solved Examples (Easy → Hard)

1

Evaluate $\int (5x - 3)^7 \, dx$.

Easy

Solution

Step 1 — Choose linear substitution: Let $y = 5x - 3$.

Step 2 — Find the differential: Differentiate both sides to find $dx$ in terms of $dy$:
$$dy = 5 \, dx \implies dx = \frac{dy}{5}$$

Step 3 — Substitute and integrate: Swap the variables inside the integral:
$$\int y^7 \left(\frac{dy}{5}\right) = \frac{1}{5} \int y^7 \, dy$$
Apply the power rule:
$$= \frac{1}{5} \cdot \frac{y^8}{8} + C = \frac{1}{40}y^8 + C$$

Step 4 — Substitute back: Replace $y$ with its original term $5x - 3$:
$$= \frac{1}{40}(5x - 3)^8 + C$$

∴ \frac{1}{40}(5x - 3)^8 + C

2

Evaluate $\int \frac{(\ln x)^2}{x} \, dx$.

Medium

Solution

Step 1 — Identify the derivative pattern: Notice that $\frac{1}{x}$ is the derivative of $\ln x$. This suggests setting up a substitution. Let $y = \ln x$.

Step 2 — Differentiate to find dy:
$$dy = \frac{1}{x} \, dx$$

Step 3 — Substitute the variables: Replace $\ln x$ with $y$ and $\frac{1}{x} \, dx$ with $dy$:
$$\int y^2 \, dy$$

Step 4 — Integrate and back-substitute: Apply the power rule and replace $y$ with $\ln x$:
$$= \frac{y^3}{3} + C = \frac{(\ln x)^3}{3} + C$$

∴ \frac{(\ln x)^3}{3} + C

3

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

Hard

Solution

Step 1 — Identify product integration technique: This is a product of two different function types. We use Integration by Parts. According to ILATE, the algebraic term ($x^2$) is first, and the trigonometric term ($\sin x$) is second. Set:
$$u = x^2, \quad dv = \sin x \, dx$$
Find the derivative and integral of these choices:
$$du = 2x \, dx, \quad v = -\cos x$$

Step 2 — Apply the formula: Use the Integration by Parts rule:
$$\int u \, dv = u v - \int v \, du$$
$$I = x^2(-\cos x) - \int (-\cos x)(2x \, dx) = -x^2\cos x + 2\int x\cos x \, dx$$

Step 3 — Apply parts a second time: We still have a product in $\int x\cos x \, dx$. Apply parts again with:
$$u = x, \quad dv = \cos x \, dx$$
$$du = dx, \quad v = \sin x$$
Calculate the inner integral:
$$\int x\cos x \, dx = x\sin x - \int \sin x \, dx = x\sin x - (-\cos x) = x\sin x + \cos x$$

Step 4 — Combine the results: Combine both steps to form the final expression, keeping track of constant coefficients:
$$I = -x^2\cos x + 2(x\sin x + \cos x) + C$$
$$= -x^2\cos x + 2x\sin x + 2\cos x + C$$

∴ -x^2\cos x + 2x\sin x + 2\cos x + C

MCQ Practice

Mixed Level — MCQ

1

What is the integral of $e^{4x}$?

Correct!

Incorrect. Correct: $\frac{1}{4}e^{4x} + C$

Solution

For the integral $\int e^{ax} \, dx$, the result is $\frac{e^{ax}}{a} + C$.
Substituting $a = 4$:
$$\int e^{4x} \, dx = \frac{1}{4}e^{4x} + C$$

∴ $\frac{1}{4}e^{4x} + C$

2

To evaluate $\int \frac{1}{x^2+25} dx$, the best substitution is:

Correct!

Incorrect. Correct: $x = 5 \tan \theta$

Solution

For the integrand involving $x^2 + a^2$ where $a^2 = 25$ ($a = 5$), the trigonometric substitution $x = a\tan\theta$ simplifies the expression:
$$x = 5\tan\theta$$

∴ $x = 5 \tan \theta$

3

In $\int x \ln x \, dx$, which function is chosen as '$u$' for Integration by Parts?

Correct!

Incorrect. Correct: $\ln x$

Solution

According to the ILATE priority rule:
1. Inverse Trig
2. Logarithmic ($\ln x$)
3. Algebraic ($x$)
Therefore, $u = \ln x$.

∴ $\ln x$

4

The value of $\int \sec^2(3x+2) dx$ is:

Correct!

Incorrect. Correct: $\frac{1}{3} \tan(3x+2) + C$

Solution

Applying the formula $\int \sec^2(ax+b) \, dx = \frac{1}{a}\tan(ax+b) + C$:
With $a=3$ and $b=2$, we get:
$$\int \sec^2(3x+2) \, dx = \frac{1}{3}\tan(3x+2) + C$$

∴ $\frac{1}{3} \tan(3x+2) + C$

5

What is $\int \frac{f'(x)}{f(x)} dx$?

Correct!

Incorrect. Correct: $\ln|f(x)| + C$

Solution

Using substitution, let $y = f(x) \implies dy = f'(x) \, dx$.
$$\int \frac{f'(x)}{f(x)} \, dx = \int \frac{dy}{y} = \ln|y| + C = \ln|f(x)| + C$$

∴ $\ln|f(x)| + C$

6

$\int \frac{dx}{\sqrt{9-x^2}}$ equals:

Correct!

Incorrect. Correct: $\sin^{-1}(x/3) + C$

Solution

Using the standard formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$:
With $a^2 = 9$ ($a = 3$):
$$\int \frac{dx}{\sqrt{9-x^2}} = \sin^{-1}\left(\frac{x}{3}\right) + C$$

∴ $\sin^{-1}(x/3) + C$

7

The derivative of the first function in Integration by Parts is:

Correct!

Incorrect. Correct: Differentiated

Solution

In the Integration by Parts formula:
$$\int u \, dv = u v - \int v \, du$$
The term $du$ represents the derivative of $u$ ($du = u' \, dx$). Thus, the first function is differentiated.

∴ Differentiated

8

Evaluate $\int \frac{dx}{x \ln x}$:

Correct!

Incorrect. Correct: $\ln|\ln x| + C$

Solution

Let $y = \ln x \implies dy = \frac{1}{x} \, dx$.
The integral simplifies to:
$$\int \frac{1}{y} \, dy = \ln|y| + C = \ln|\ln x| + C$$

∴ $\ln|\ln x| + C$

9

$\int e^x ( \sin x + \cos x ) dx$ is:

Correct!

Incorrect. Correct: $e^x \sin x + C$

Solution

Using the exponential integral theorem:
$$\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$$
Here, $f(x) = \sin x$ and $f'(x) = \cos x$. Thus:
$$\int e^x (\sin x + \cos x) \, dx = e^x \sin x + C$$

∴ $e^x \sin x + C$

10

If $y = x^2$, then $dy$ is:

Correct!

Incorrect. Correct: $2x \, dx$

Solution

Since $y = x^2$, taking the derivative gives:
$$\frac{dy}{dx} = 2x$$
Multiplying both sides by the differential $dx$ gives:
$$dy = 2x \, dx$$

∴ $2x \, dx$

Short Questions (2 Marks Each)

1

Evaluate $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$.

[2 marks]

Solution

Step 1 — Substitute the root: Let $y = \sqrt{x} = x^{1/2}$.

Step 2 — Find the differential:
$$dy = \frac{1}{2\sqrt{x}} \, dx \implies 2 \, dy = \frac{1}{\sqrt{x}} \, dx$$

Step 3 — Substitute and integrate:
$$\int e^y (2 \, dy) = 2\int e^y \, dy = 2e^y + C$$
Replace $y$ with $\sqrt{x}$:
$$= 2e^{\sqrt{x}} + C$$

∴ 2e^{\sqrt{x}} + C

2

Find $\int \tan x \, dx$.

[2 marks]

Solution

Step 1 — Rewrite in terms of sine and cosine:
$$\int \frac{\sin x}{\cos x} \, dx$$

Step 2 — Use substitution: Let $y = \cos x \implies dy = -\sin x \, dx \implies -dy = \sin x \, dx$.

Step 3 — Substitute and integrate:
$$\int \frac{-dy}{y} = -\ln|y| + C = -\ln|\cos x| + C$$
Using logarithmic properties:
$$= \ln|(\cos x)^{-1}| + C = \ln|\sec x| + C$$

∴ \ln|\sec x| + C

3

Evaluate $\int x e^{x^2} dx$.

[2 marks]

Solution

Step 1 — Substitution selection: Let $y = x^2$.

Step 2 — Find differential:
$$dy = 2x \, dx \implies \frac{dy}{2} = x \, dx$$

Step 3 — Substitute and integrate:
$$\int e^y \left(\frac{dy}{2}\right) = \frac{1}{2}\int e^y \, dy = \frac{1}{2}e^y + C$$
Replace $y$ with $x^2$:
$$= \frac{1}{2}e^{x^2} + C$$

∴ \frac{1}{2}e^{x^2} + C

4

Find $\int \frac{x}{1+x^4} dx$.

[2 marks]

Solution

Step 1 — Rewrite the denominator: Express $x^4$ as $(x^2)^2$:
$$\int \frac{x}{1+(x^2)^2} \, dx$$

Step 2 — Substitution: Let $y = x^2 \implies dy = 2x \, dx \implies \frac{dy}{2} = x \, dx$.

Step 3 — Substitute and integrate:
$$\frac{1}{2}\int \frac{dy}{1+y^2} = \frac{1}{2}\tan^{-1}(y) + C$$
Replace $y$ with $x^2$:
$$= \frac{1}{2}\tan^{-1}(x^2) + C$$

∴ \frac{1}{2}\tan^{-1}(x^2) + C

5

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

[2 marks]

Solution

Step 1 — Set up Integration by Parts: Treat $\ln x$ as a product with $1$:
$$\int (\ln x \cdot 1) \, dx$$
By ILATE, let $u = \ln x$ and $dv = 1 \, dx$.
$$du = \frac{1}{x} \, dx, \quad v = x$$

Step 2 — Apply the formula:
$$I = u v - \int v \, du = x\ln x - \int x \left(\frac{1}{x}\right) dx$$
$$= x\ln x - \int 1 \, dx = x\ln x - x + C$$

∴ x\ln x - x + C

6

Evaluate $\int \frac{dx}{\sqrt{1-4x^2}}$.

[2 marks]

Solution

Step 1 — Format the expression: Express $4x^2$ as $(2x)^2$:
$$\int \frac{dx}{\sqrt{1-(2x)^2}}$$

Step 2 — Substitution: Let $2x = \sin\theta \implies 2 \, dx = \cos\theta \, d\theta \implies dx = \frac{1}{2}\cos\theta \, d\theta$.
Also, $\sqrt{1-(2x)^2} = \sqrt{1-\sin^2\theta} = \cos\theta$.

Step 3 — Substitute and integrate:
$$\int \frac{\frac{1}{2}\cos\theta \, d\theta}{\cos\theta} = \frac{1}{2}\int 1 \, d\theta = \frac{1}{2}\theta + C$$
Since $2x = \sin\theta \implies \theta = \sin^{-1}(2x)$:
$$= \frac{1}{2}\sin^{-1}(2x) + C$$

∴ \frac{1}{2}\sin^{-1}(2x) + C

7

Find $\int x \cos(x^2+5) dx$.

[2 marks]

Solution

Step 1 — Choose substitution: Let $y = x^2+5$.

Step 2 — Differentiate:
$$dy = 2x \, dx \implies \frac{dy}{2} = x \, dx$$

Step 3 — Substitute and integrate:
$$\int \cos y \left(\frac{dy}{2}\right) = \frac{1}{2}\int \cos y \, dy = \frac{1}{2}\sin y + C$$
Replace $y$ with $x^2+5$:
$$= \frac{1}{2}\sin(x^2+5) + C$$

∴ \frac{1}{2}\sin(x^2+5) + C

8

Evaluate $\int \frac{1}{x(\ln x)^2} dx$.

[2 marks]

Solution

Step 1 — Choose substitution: Let $y = \ln x$.

Step 2 — Differentiate:
$$dy = \frac{1}{x} \, dx$$

Step 3 — Substitute and integrate:
$$\int \frac{dy}{y^2} = \int y^{-2} \, dy = \frac{y^{-1}}{-1} + C = -\frac{1}{y} + C$$
Replace $y$ with $\ln x$:
$$= -\frac{1}{\ln x} + C$$

∴ -\frac{1}{\ln x} + C

9

Find $\int \frac{\sin(\tan^{-1} x)}{1+x^2} dx$.

[2 marks]

Solution

Step 1 — Choose substitution: Let $y = \tan^{-1} x$.

Step 2 — Differentiate:
$$dy = \frac{1}{1+x^2} \, dx$$

Step 3 — Substitute and integrate:
$$\int \sin y \, dy = -\cos y + C$$
Replace $y$ with $\tan^{-1} x$:
$$= -\cos(\tan^{-1} x) + C$$

∴ -\cos(\tan^{-1} x) + C

10

Evaluate $\int e^x \sec^2(e^x) dx$.

[2 marks]

Solution

Step 1 — Choose substitution: Let $y = e^x$.

Step 2 — Differentiate:
$$dy = e^x \, dx$$

Step 3 — Substitute and integrate:
$$\int \sec^2 y \, dy = \tan y + C$$
Replace $y$ with $e^x$:
$$= \tan(e^x) + C$$

∴ \tan(e^x) + C

Long Questions (4 Marks Each)

1

Evaluate $\int \sqrt{a^2-x^2} \, dx$.

[4 marks]

Solution

Step 1 — Use trigonometric substitution: Let $x = a\sin\theta \implies dx = a\cos\theta \, d\theta$.
Also, $\sqrt{a^2-x^2} = \sqrt{a^2(1-\sin^2\theta)} = a\cos\theta$.

Step 2 — Substitute into the integral:
$$\int a\cos\theta \cdot a\cos\theta \, d\theta = a^2 \int \cos^2\theta \, d\theta$$

Step 3 — Apply power-reduction identity and integrate:
$$= \frac{a^2}{2} \int (1+\cos 2\theta) d\theta = \frac{a^2}{2}\left[\theta + \frac{\sin 2\theta}{2}\right] + C$$
Rewrite $\sin 2\theta$ as $2\sin\theta\cos\theta$ to ease back-substitution:
$$= \frac{a^2}{2}\left[\theta + \sin\theta\cos\theta\right] + C$$

Step 4 — Perform back-substitution: Since $\sin\theta = \frac{x}{a}$, we have $\theta = \sin^{-1}\left(\frac{x}{a}\right)$ and $\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt{a^2-x^2}}{a}$:
$$= \frac{a^2}{2}\left[\sin^{-1}\left(\frac{x}{a}\right) + \frac{x}{a} \cdot \frac{\sqrt{a^2-x^2}}{a}\right] + C$$
$$= \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + \frac{x\sqrt{a^2-x^2}}{2} + C$$

∴ \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + \frac{x\sqrt{a^2-x^2}}{2} + C

2

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

[4 marks]

Solution

Step 1 — Define the integral and apply Integration by Parts: Let $I = \int e^x \cos x \, dx$.
Using ILATE, Algebraic/Trig/Exponential priority sets: $u = \cos x, dv = e^x \, dx \implies du = -\sin x \, dx, v = e^x$.
$$I = e^x\cos x - \int e^x(-\sin x) \, dx = e^x\cos x + \int e^x\sin x \, dx$$

Step 2 — Apply parts a second time: We still have a product in the second term. Let $u = \sin x, dv = e^x \, dx \implies du = \cos x \, dx, v = e^x$.
$$\int e^x\sin x \, dx = e^x\sin x - \int e^x\cos x \, dx = e^x\sin x - I$$

Step 3 — Set up the equation and solve for I: Substitute this back into our primary equation:
$$I = e^x\cos x + e^x\sin x - I$$
$$2I = e^x(\sin x + \cos x)$$
$$I = \frac{e^x}{2}(\sin x + \cos x) + C$$

∴ \frac{e^x}{2}(\sin x + \cos x) + C

3

Evaluate $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$.

[4 marks]

Solution

Step 1 — Set up trigonometric substitution: Let $x = \sin\theta \implies dx = \cos\theta \, d\theta$.
The denominator becomes $\sqrt{1-\sin^2\theta} = \cos\theta$.

Step 2 — Substitute into the integral:
$$\int \frac{\sin^2\theta}{\cos\theta} \cos\theta \, d\theta = \int \sin^2\theta \, d\theta$$

Step 3 — Apply power-reduction identity and integrate:
$$= \frac{1}{2} \int (1-\cos 2\theta) d\theta = \frac{1}{2}\left[\theta - \frac{\sin 2\theta}{2}\right] + C = \frac{1}{2}\left[\theta - \sin\theta\cos\theta\right] + C$$

Step 4 — Perform back-substitution: Since $x = \sin\theta \implies \theta = \sin^{-1} x$ and $\cos\theta = \sqrt{1-x^2}$:
$$= \frac{1}{2}\sin^{-1} x - \frac{1}{2}x\sqrt{1-x^2} + C$$

∴ \frac{1}{2}\sin^{-1} x - \frac{1}{2}x\sqrt{1-x^2} + C

4

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

[4 marks]

Solution

Step 1 — Split the term and apply Integration by Parts: Let $I = \int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx$.
Set $u = \sec x$ and $dv = \sec^2 x \, dx \implies du = \sec x\tan x \, dx, v = \tan x$.
$$I = \sec x\tan x - \int \tan x(\sec x\tan x) \, dx = \sec x\tan x - \int \sec x\tan^2 x \, dx$$

Step 2 — Apply trigonometric identity: Replace $\tan^2 x$ with $\sec^2 x - 1$:
$$I = \sec x\tan x - \int \sec x(\sec^2 x - 1) \, dx$$
$$I = \sec x\tan x - \int \sec^3 x \, dx + \int \sec x \, dx$$

Step 3 — Express in terms of I and integrate:
$$I = \sec x\tan x - I + \ln|\sec x + \tan x|$$
$$2I = \sec x\tan x + \ln|\sec x + \tan x|$$
$$I = \frac{1}{2}\left[\sec x\tan x + \ln|\sec x + \tan x|\right] + C$$

∴ \frac{1}{2}\left[\sec x\tan x + \ln|\sec x + \tan x|\right] + C

5

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

[4 marks]

Solution

Step 1 — Set up parts selection: By ILATE, Logarithmic ($\ln x$) is $u$, and Algebraic ($x^2$) is $dv$.
$$u = \ln x, \quad dv = x^2 \, dx \implies du = \frac{1}{x} \, dx, \quad v = \frac{x^3}{3}$$

Step 2 — Apply the formula:
$$I = u v - \int v \, du = \ln x \left(\frac{x^3}{3}\right) - \int \frac{x^3}{3} \left(\frac{1}{x}\right) dx$$
$$= \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2 \, dx$$

Step 3 — Integrate the remaining term:
$$= \frac{x^3}{3}\ln x - \frac{1}{3}\left(\frac{x^3}{3}\right) + C = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$$

∴ \frac{x^3}{3}\ln x - \frac{x^3}{9} + C

6

Evaluate $\int \frac{e^x+1}{e^x-1} \, dx$.

[4 marks]

Solution

Step 1 — Divide numerator and denominator by $e^{x/2}$ to balance terms:
$$\int \frac{e^{x/2} + e^{-x/2}}{e^{x/2} - e^{-x/2}} \, dx$$

Step 2 — Use substitution: Let $y = e^{x/2} - e^{-x/2}$. Differentiating gives:
$$dy = \frac{1}{2}\left(e^{x/2} + e^{-x/2}\right) dx \implies 2 \, dy = \left(e^{x/2} + e^{-x/2}\right) dx$$

Step 3 — Substitute and integrate:
$$\int \frac{2 \, dy}{y} = 2\int \frac{1}{y} \, dy = 2\ln|y| + C$$
Replace $y$ with $e^{x/2} - e^{-x/2}$:
$$= 2\ln\left|e^{x/2} - e^{-x/2}\right| + C$$

∴ 2\ln\left|e^{x/2} - e^{-x/2}\right| + C

7

Evaluate $\int \tan^4 \theta \, d\theta$.

[4 marks]

Solution

Step 1 — Split the power:
$$\int \tan^2\theta \cdot \tan^2\theta \, d\theta$$
Replace the second term with $\sec^2\theta - 1$:
$$\int \tan^2\theta(\sec^2\theta - 1) \, d\theta = \int \tan^2\theta\sec^2\theta \, d\theta - \int \tan^2\theta \, d\theta$$

Step 2 — Solve both integrals separately:
• For the first integral $\int \tan^2\theta\sec^2\theta \, d\theta$: let $y = \tan\theta \implies dy = \sec^2\theta \, d\theta$. The expression simplifies to $\int y^2 \, dy = \frac{y^3}{3} = \frac{\tan^3\theta}{3}$.
• For the second integral $\int \tan^2\theta \, d\theta$: rewrite as $\int (\sec^2\theta - 1) \, d\theta = \tan\theta - \theta$.

Step 3 — Combine the results:
$$= \frac{\tan^3\theta}{3} - (\tan\theta - \theta) + C = \frac{\tan^3\theta}{3} - \tan\theta + \theta + C$$

∴ \frac{\tan^3\theta}{3} - \tan\theta + \theta + C

8

Evaluate $\int \frac{x^3}{\sqrt{a^2-x^2}} \, dx$.

[4 marks]

Solution

Step 1 — Set up trigonometric substitution: Let $x = a\sin\theta \implies dx = a\cos\theta \, d\theta$.
The denominator becomes $\sqrt{a^2-a^2\sin^2\theta} = a\cos\theta$.

Step 2 — Substitute into the integral:
$$\int \frac{a^3\sin^3\theta}{a\cos\theta} a\cos\theta \, d\theta = a^3\int \sin^3\theta \, d\theta$$

Step 3 — Integrate the sine cubic function: Rewrite $\sin^3\theta$ as $\sin\theta(1-\cos^2\theta)$:
$$a^3\int \sin\theta(1-\cos^2\theta) \, d\theta$$
Let $u = \cos\theta \implies du = -\sin\theta \, d\theta \implies -du = \sin\theta \, d\theta$.
$$= a^3\int -(1-u^2) \, du = a^3\int (u^2-1) \, du = a^3\left[\frac{u^3}{3} - u\right] + C = a^3\left[\frac{\cos^3\theta}{3} - \cos\theta\right] + C$$

Step 4 — Perform back-substitution: Since $\sin\theta = \frac{x}{a} \implies \cos\theta = \sqrt{1-\frac{x^2}{a^2}} = \frac{\sqrt{a^2-x^2}}{a}$:
$$= a^3\left[ \frac{(a^2-x^2)^{3/2}}{3a^3} - \frac{\sqrt{a^2-x^2}}{a} \right] + C = \frac{1}{3}(a^2-x^2)^{3/2} - a^2\sqrt{a^2-x^2} + C$$

∴ \frac{1}{3}(a^2-x^2)^{3/2} - a^2\sqrt{a^2-x^2} + C

9

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

[4 marks]

Solution

Step 1 — Apply Integration by Parts: Treat the function as a product with $1$:
$$\int (\sin^{-1} x \cdot 1) \, dx$$
Set $u = \sin^{-1} x$ and $dv = 1 \, dx \implies du = \frac{1}{\sqrt{1-x^2}} \, dx, v = x$.

Step 2 — Substitute into the parts formula:
$$I = x\sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} \, dx$$

Step 3 — Solve the remaining algebraic integral: For $\int \frac{x}{\sqrt{1-x^2}} \, dx$, use substitution. Let $y = 1-x^2 \implies dy = -2x \, dx \implies -\frac{dy}{2} = x \, dx$.
$$\int y^{-1/2} \left(-\frac{dy}{2}\right) = -\frac{1}{2} \cdot \frac{y^{1/2}}{1/2} = -\sqrt{y} = -\sqrt{1-x^2}$$

Step 4 — Combine terms:
$$I = x\sin^{-1} x - \left(-\sqrt{1-x^2}\right) + C = x\sin^{-1} x + \sqrt{1-x^2} + C$$

∴ x\sin^{-1} x + \sqrt{1-x^2} + C

10

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

[4 marks]

Solution

Step 1 — Set up parts selection: By ILATE, $u = x^2$ (algebraic) and $dv = e^{2x} \, dx$ (exponential).
$$du = 2x \, dx, \quad v = \frac{e^{2x}}{2}$$
$$I = x^2\left(\frac{e^{2x}}{2} ight) - \int \left(\frac{e^{2x}}{2}\right) (2x \, dx) = \frac{x^2 e^{2x}}{2} - \int x e^{2x} \, dx$$

Step 2 — Apply parts a second time: For $\int x e^{2x} \, dx$, let $u = x$ and $dv = e^{2x} \, dx \implies du = dx, v = \frac{e^{2x}}{2}$.
$$\int x e^{2x} \, dx = x\left(\frac{e^{2x}}{2} ight) - \int \frac{e^{2x}}{2} \, dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4}$$

Step 3 — Combine and simplify the final equation:
$$I = \frac{x^2 e^{2x}}{2} - \left( \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} \right) + C$$
$$= e^{2x} \left[ \frac{x^2}{2} - \frac{x}{2} + \frac{1}{4} \right] + C$$

∴ e^{2x} \left[ \frac{x^2}{2} - \frac{x}{2} + \frac{1}{4} \right] + C

Chapter Test — Full Mixed Paper

Attempt in 45 minutes without notes. This mixed diagnostic paper consists of sections aligning with NEB board examinations.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

$\int \frac{1}{x} \, dx$ is:

Correct!

Incorrect. Correct: $\ln x$

Solution

By the standard reciprocal power integral rule:
$$\int \frac{1}{x} \, dx = \ln|x| + C$$

∴ $\ln x$

2

The correct trigonometric substitution for $\sqrt{x^2-9}$ is:

Correct!

Incorrect. Correct: $3\sec\theta$

Solution

For the radical form $\sqrt{x^2 - a^2}$ with $a = 3$, the best substitution is:
$$x = 3\sec\theta$$

∴ $3\sec\theta$

3

In the ILATE hierarchy, 'E' stands for:

Correct!

Incorrect. Correct: Exponential

Solution

ILATE is an acronym representing function priorities in product integration:
• I: Inverse trigonometric
• L: Logarithmic
• A: Algebraic
• T: Trigonometric
• E: Exponential

∴ Exponential

2 Section B — Short Questions (2 Marks Each)

1

Evaluate $\int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx$.

[2 marks]

Solution

Let $y = \sqrt{x} \implies dy = \frac{1}{2\sqrt{x}} \, dx \implies 2 \, dy = \frac{dx}{\sqrt{x}}$.
$$\int \sin y (2 \, dy) = 2 \int \sin y \, dy = -2\cos y + C = -2\cos\sqrt{x} + C$$

∴ -2\cos\sqrt{x} + C

2

Evaluate $\int x \cos x \, dx$.

[2 marks]

Solution

Use Integration by Parts with $u = x$ and $dv = \cos x \, dx \implies du = dx, v = \sin x$.
$$I = x\sin x - \int \sin x \, dx = x\sin x - (-\cos x) + C = x\sin x + \cos x + C$$

∴ x\sin x + \cos x + C

3

Find $\int \frac{e^x}{1+e^x} \, dx$.

[2 marks]

Solution

Let $y = 1+e^x \implies dy = e^x \, dx$.
$$\int \frac{dy}{y} = \ln|y| + C = \ln\left|1+e^x\right| + C$$

∴ \ln\left|1+e^x\right| + C

3 Section C — Long Questions (8 Marks Each)

1

Evaluate $\int e^{3x} \sin 2x \, dx$.

[8 marks]

Solution

Let $I = \int e^{3x}\sin 2x \, dx$. Use parts with $u = \sin 2x, dv = e^{3x} \, dx \implies du = 2\cos 2x \, dx, v = \frac{e^{3x}}{3}$.
$$I = \frac{e^{3x}\sin 2x}{3} - \frac{2}{3}\int e^{3x}\cos 2x \, dx$$
Apply parts again to the remaining integral with $u = \cos 2x, dv = e^{3x} \, dx \implies du = -2\sin 2x \, dx, v = \frac{e^{3x}}{3}$:
$$\int e^{3x}\cos 2x \, dx = \frac{e^{3x}\cos 2x}{3} - \int \frac{e^{3x}}{3}(-2\sin 2x) dx = \frac{e^{3x}\cos 2x}{3} + \frac{2}{3}I$$
Combine both steps:
$$I = \frac{e^{3x}\sin 2x}{3} - \frac{2}{3}\left( \frac{e^{3x}\cos 2x}{3} + \frac{2}{3}I \right)$$
$$I = \frac{e^{3x}\sin 2x}{3} - \frac{2e^{3x}\cos 2x}{9} - \frac{4}{9}I$$
$$\frac{13}{9}I = e^{3x} \left[ \frac{3\sin 2x - 2\cos 2x}{9} \right]$$
$$I = \frac{e^{3x}(3\sin 2x - 2\cos 2x)}{13} + C$$

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

2

Evaluate $\int \frac{dx}{x^2 \sqrt{1+x^2}}$.

[8 marks]

Solution

Let $x = \tan\theta \implies dx = \sec^2\theta \, d\theta$.
The denominator becomes $x^2\sqrt{1+x^2} = \tan^2\theta\sqrt{1+\tan^2\theta} = \tan^2\theta\sec\theta$.
$$\int \frac{\sec^2\theta \, d\theta}{\tan^2\theta\sec\theta} = \int \frac{\sec\theta}{\tan^2\theta} \, d\theta$$
Convert to sine and cosine:
$$= \int \frac{1}{\cos\theta} \cdot \frac{\cos^2\theta}{\sin^2\theta} \, d\theta = \int \frac{\cos\theta}{\sin^2\theta} \, d\theta$$
Let $u = \sin\theta \implies du = \cos\theta \, d\theta$.
$$= \int \frac{du}{u^2} = -\frac{1}{u} + C = -\csc\theta + C$$
Since $x = \tan\theta$, we have a right-angled triangle where opposite = $x$, adjacent = $1$, and hypotenuse = $\sqrt{1+x^2}$. Therefore, $\csc\theta = \frac{\sqrt{1+x^2}}{x}$:
$$= -\frac{\sqrt{1+x^2}}{x} + C$$

∴ -\frac{\sqrt{1+x^2}}{x} + C

One-Page Revision Cheat Sheet

1

Antiderivative: The operations of differentiation and integration are mathematical inverses.

2

Substitution Method: Select $y = f(x)$ such that the derivative $f'(x) \, dx$ matches another active component in your integrand.

3

Trig Substitutions: Let $x = a\sin\theta$ for $\sqrt{a^2-x^2}$, let $x = a\tan\theta$ for $\sqrt{a^2+x^2}$, and let $x = a\sec\theta$ for $\sqrt{x^2-a^2}$.

4

Integration by Parts: Use the pattern $\int u \, dv = u v - \int v \, du$. Prioritize term selection for $u$ using the ILATE order.

5

Product priority (ILATE): Inverse Trig (I) > Logarithmic (L) > Algebraic (A) > Trigonometric (T) > Exponential (E).

6

Self-Check Strategy: Take the derivative of your final expression. It must match the original function inside your integral to prove correctness.