The Hook: Reversing a Speedometer
Imagine you’re watching the car chase scene from Fast X (2023). The speedometer shows the car’s speed changing constantly—sometimes 60 km/h, sometimes 100 km/h. Now here’s the question:
If someone gives you only the speedometer readings (speed at every instant), can you figure out the total distance traveled?
This is exactly what integration does! If differentiation gives you the rate of change (speed from distance), integration reverses it to get the accumulated quantity (distance from speed).
In JEE, this concept appears in 3-4 questions every year—making it a high-yield topic.
The Core Concept
What is an Indefinite Integral?
An indefinite integral (or antiderivative) is the reverse operation of differentiation.
$$\boxed{\int f(x) \, dx = F(x) + C}$$where:
- $\int$ is the integration symbol (elongated S for “sum”)
- $f(x)$ is the integrand (function to integrate)
- $dx$ indicates integration with respect to $x$
- $F(x)$ is the antiderivative such that $\frac{d}{dx}F(x) = f(x)$
- $C$ is the constant of integration (crucial!)
In simple terms: Integration asks: “Which function, when differentiated, gives me $f(x)$?”
Why the “+C”?
When you differentiate $x^2$, you get $2x$. When you differentiate $x^2 + 5$, you still get $2x$. When you differentiate $x^2 - 100$, you still get $2x$!
The derivative “kills” the constant. So when reversing (integrating), we must add back an arbitrary constant $C$ because we don’t know what was lost.
JEE Trap: Forgetting $+C$ in indefinite integrals costs you marks!
Interactive Demo: Visualize Antiderivatives
Plot a function like $f(x) = 2x$ and see how its antiderivative $F(x) = x^2 + C$ changes with different values of C. This shows why we need the constant of integration!
Standard Integral Formulas
Power Functions
$$\boxed{\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1}$$Memory Trick: “Increase Power, Divide” → IPD Rule
Special case:
$$\boxed{\int x^{-1} \, dx = \int \frac{1}{x} \, dx = \ln|x| + C}$$Exponential Functions
$$\boxed{\int e^x \, dx = e^x + C}$$ $$\boxed{\int a^x \, dx = \frac{a^x}{\ln a} + C, \quad a > 0, a \neq 1}$$Pattern: $e^x$ is special—it’s its own derivative AND its own integral!
Trigonometric Functions
| Function | Integral |
|---|---|
| $\int \sin x \, dx$ | $-\cos x + C$ |
| $\int \cos x \, dx$ | $\sin x + C$ |
| $\int \sec^2 x \, dx$ | $\tan x + C$ |
| $\int \csc^2 x \, dx$ | $-\cot x + C$ |
| $\int \sec x \tan x \, dx$ | $\sec x + C$ |
| $\int \csc x \cot x \, dx$ | $-\csc x + C$ |
Memory Trick: “Sign Changes for Sin Cos” → Sine’s integral has a negative sign!
Inverse Trigonometric Functions
$$\boxed{\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x + C}$$ $$\boxed{\int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C}$$ $$\boxed{\int \frac{1}{x\sqrt{x^2-1}} \, dx = \sec^{-1} x + C}$$Properties of Indefinite Integrals
Linearity Properties
Constant Multiple Rule:
$$\int k \cdot f(x) \, dx = k \int f(x) \, dx$$Sum/Difference Rule:
$$\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx$$
Problem: Find $\int (3x^2 - 5\cos x + 2e^x) \, dx$
Solution:
$$\begin{aligned} \int (3x^2 - 5\cos x + 2e^x) \, dx &= 3\int x^2 \, dx - 5\int \cos x \, dx + 2\int e^x \, dx \\ &= 3 \cdot \frac{x^3}{3} - 5\sin x + 2e^x + C \\ &= x^3 - 5\sin x + 2e^x + C \end{aligned}$$Non-Linear Operations (Important!)
WRONG: $\int [f(x) \cdot g(x)] \, dx \neq \int f(x) \, dx \cdot \int g(x) \, dx$
WRONG: $\int \frac{f(x)}{g(x)} \, dx \neq \frac{\int f(x) \, dx}{\int g(x) \, dx}$
Example: $\int x \cdot e^x \, dx \neq \frac{x^2}{2} \cdot e^x$ (This needs integration by parts!)
Fundamental Integration Techniques
Technique 1: Direct Formula Application
Problem: Evaluate $\int \left(x^4 + \frac{3}{x^2} - \sqrt{x}\right) dx$
Solution:
$$\begin{aligned} \int \left(x^4 + 3x^{-2} - x^{1/2}\right) dx &= \frac{x^5}{5} + 3 \cdot \frac{x^{-1}}{-1} - \frac{x^{3/2}}{3/2} + C \\ &= \frac{x^5}{5} - \frac{3}{x} - \frac{2x^{3/2}}{3} + C \end{aligned}$$Technique 2: Simplification Before Integration
Problem: Find $\int \frac{x^3 + 3x^2 + 3x + 1}{x^2} \, dx$
Solution: First, simplify by dividing:
$$\frac{x^3 + 3x^2 + 3x + 1}{x^2} = x + 3 + \frac{3}{x} + \frac{1}{x^2}$$Now integrate:
$$\begin{aligned} \int \left(x + 3 + 3x^{-1} + x^{-2}\right) dx &= \frac{x^2}{2} + 3x + 3\ln|x| - \frac{1}{x} + C \end{aligned}$$Technique 3: Trigonometric Identities
Problem: Evaluate $\int \sin^2 x \, dx$
Solution: Use the identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$
$$\begin{aligned} \int \sin^2 x \, dx &= \int \frac{1 - \cos 2x}{2} \, dx \\ &= \frac{1}{2}\int 1 \, dx - \frac{1}{2}\int \cos 2x \, dx \\ &= \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin 2x}{2} + C \\ &= \frac{x}{2} - \frac{\sin 2x}{4} + C \end{aligned}$$Common Algebraic Manipulations
Rationalization Technique
Problem: Find $\int \frac{1}{x + \sqrt{x}} \, dx$
Solution: Rationalize by multiplying by $\frac{\sqrt{x}}{\sqrt{x}}$:
$$\frac{1}{x + \sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x} + 1)} = \frac{1}{\sqrt{x}(\sqrt{x} + 1)} \cdot \frac{\sqrt{x} - 1}{\sqrt{x} - 1}$$Actually, simpler approach: Factor out $\sqrt{x}$:
$$\frac{1}{x + \sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x} + 1)}$$Let $u = \sqrt{x} + 1$ (we’ll learn substitution in the next topic!)
For now, recognize: This splits into partial fractions.
Memory Tricks & Patterns
Master Formula Sheet
Powers:
- $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (IPD Rule)
- $\int \frac{1}{x} \, dx = \ln|x| + C$
Exponentials: 3. $\int e^x \, dx = e^x + C$ 4. $\int a^x \, dx = \frac{a^x}{\ln a} + C$
Trigonometric (The Big 6): 5. $\int \sin x \, dx = -\cos x + C$ (negative!) 6. $\int \cos x \, dx = \sin x + C$ (positive!) 7. $\int \sec^2 x \, dx = \tan x + C$ 8. $\int \csc^2 x \, dx = -\cot x + C$ 9. $\int \sec x \tan x \, dx = \sec x + C$ 10. $\int \csc x \cot x \, dx = -\csc x + C$
Inverse Trigonometric: 11. $\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x + C$ 12. $\int \frac{-1}{\sqrt{1-x^2}} \, dx = \cos^{-1} x + C$ 13. $\int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C$ 14. $\int \frac{-1}{1+x^2} \, dx = \cot^{-1} x + C$ 15. $\int \frac{1}{|x|\sqrt{x^2-1}} \, dx = \sec^{-1} x + C$
Memorize these 15—they appear in 90% of JEE integration problems!
Sign Pattern for Trig Functions
Mnemonic: “Some Cops Have Caught Students Cutting”
- Some = $\sin x$ integrates to negative ($-\cos x$)
- Cops = $\cos x$ integrates to positive ($\sin x$)
- Have = (skip)
- Caught = $\csc^2 x$ integrates to negative ($-\cot x$)
- Students = $\sec^2 x$ integrates to positive ($\tan x$)
- Cutting = $\csc x \cot x$ integrates to negative ($-\csc x$)
When to Use Standard Formulas
Step 1: Can you simplify algebraically?
- Expand brackets
- Divide polynomials
- Use trig identities
Step 2: Does it match a standard form?
- Check the master list of 15 formulas
- Look for $\frac{f'(x)}{f(x)}$ pattern → $\ln|f(x)|$
Step 3: Need advanced technique?
- Substitution → See Integration Techniques
- By Parts → See Integration Techniques
- Partial Fractions → See Integration Techniques
Common Mistakes to Avoid
WRONG: $\int 2x \, dx = x^2$ CORRECT: $\int 2x \, dx = x^2 + C$
Why it matters: In JEE, if a problem asks for an indefinite integral, missing $+C$ is marked wrong!
WRONG: $\int x^{-1} \, dx = \frac{x^0}{0} + C$ (Undefined!) CORRECT: $\int \frac{1}{x} \, dx = \ln|x| + C$
This is a special case—memorize it separately!
WRONG: $\int \frac{1}{x} \, dx = \ln x + C$ CORRECT: $\int \frac{1}{x} \, dx = \ln|x| + C$
The domain of $\ln x$ is $x > 0$, but $\frac{1}{x}$ is defined for all $x \neq 0$. The absolute value ensures the integral works for negative $x$ too.
WRONG: $\int \sin x \, dx = \cos x + C$ CORRECT: $\int \sin x \, dx = -\cos x + C$
Check method: Differentiate your answer! $\frac{d}{dx}(-\cos x) = \sin x$ ✓
Practice Problems
Level 1: Foundation (NCERT)
Evaluate: $\int (x^3 - 2x^2 + 5x - 7) \, dx$
Solution:
$$\begin{aligned} \int (x^3 - 2x^2 + 5x - 7) \, dx &= \frac{x^4}{4} - 2 \cdot \frac{x^3}{3} + 5 \cdot \frac{x^2}{2} - 7x + C \\ &= \frac{x^4}{4} - \frac{2x^3}{3} + \frac{5x^2}{2} - 7x + C \end{aligned}$$Find: $\int (3\sin x - 4\cos x) \, dx$
Solution:
$$\begin{aligned} \int (3\sin x - 4\cos x) \, dx &= 3\int \sin x \, dx - 4\int \cos x \, dx \\ &= 3(-\cos x) - 4(\sin x) + C \\ &= -3\cos x - 4\sin x + C \end{aligned}$$Evaluate: $\int \left(e^x + \frac{1}{x} + 2^x\right) dx$
Solution:
$$\begin{aligned} \int \left(e^x + \frac{1}{x} + 2^x\right) dx &= e^x + \ln|x| + \frac{2^x}{\ln 2} + C \end{aligned}$$Level 2: JEE Main
JEE Main 2023 Pattern: Find $\int \frac{x^4 + 1}{x^2} \, dx$
Solution: Simplify first:
$$\frac{x^4 + 1}{x^2} = x^2 + x^{-2}$$ $$\begin{aligned} \int (x^2 + x^{-2}) \, dx &= \frac{x^3}{3} + \frac{x^{-1}}{-1} + C \\ &= \frac{x^3}{3} - \frac{1}{x} + C \end{aligned}$$Problem: Evaluate $\int (\sqrt{x} + \frac{1}{\sqrt{x}})^2 dx$
Solution: Expand first:
$$(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = x + 2 + \frac{1}{x}$$ $$\begin{aligned} \int \left(x + 2 + \frac{1}{x}\right) dx &= \frac{x^2}{2} + 2x + \ln|x| + C \end{aligned}$$JEE Tip: Always expand before integrating—don’t try to integrate the squared form directly!
Problem: Find $\int \frac{1 - \cos 2x}{1 + \cos 2x} \, dx$
Solution: Use half-angle formulas:
- $1 - \cos 2x = 2\sin^2 x$
- $1 + \cos 2x = 2\cos^2 x$
Now use: $\tan^2 x = \sec^2 x - 1$
$$\begin{aligned} \int \tan^2 x \, dx &= \int (\sec^2 x - 1) \, dx \\ &= \tan x - x + C \end{aligned}$$Level 3: JEE Advanced
JEE Advanced Pattern: If $\int f(x) \, dx = \psi(x)$, then find $\int x^5 f(x^3) \, dx$ in terms of $\psi$.
Solution: This requires substitution. Let $u = x^3$, then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{du}{3}$.
$$x^5 f(x^3) \, dx = x^3 \cdot x^2 f(x^3) \, dx = u \cdot f(u) \cdot \frac{du}{3}$$But we need more manipulation. Note: $x^5 = x^3 \cdot x^2 = u \cdot x^2$.
$$\int x^5 f(x^3) \, dx = \int u \cdot f(u) \cdot x^2 \, dx$$Since $u = x^3$, we have $x^2 \, dx = \frac{du}{3}$:
$$\int u \cdot f(u) \cdot \frac{du}{3} = \frac{1}{3} \int u f(u) \, du$$This still requires integration by parts or knowing more about $f$.
Key learning: Advanced problems combine multiple techniques!
Problem: Prove that $\int \frac{dx}{x^2(x^4 + 1)^{3/4}} = -\left[\frac{x^4 + 1}{x^4}\right]^{1/4} + C$
Solution: Rewrite the integrand:
$$\frac{1}{x^2(x^4 + 1)^{3/4}} = \frac{x^{-2}}{(x^4 + 1)^{3/4}}$$Divide numerator and denominator by $x^3$:
$$= \frac{x^{-2}}{x^3(x + x^{-3})^{3/4}} = \frac{1}{x^5(x + x^{-3})^{3/4}}$$Actually, better approach: Substitute $t = \frac{x^4 + 1}{x^4}$
Then $t = 1 + x^{-4}$, so $dt = -4x^{-5} \, dx$
This is complex—requires practice with advanced substitution!
For JEE Advanced: Focus on pattern recognition and creative substitutions.
Conceptual Problem: Let $I = \int \sin x \, dx$ and $J = \int \cos x \, dx$. Then which of the following is true?
A) $I^2 + J^2 = C$ (constant) B) $\frac{dI}{dx} = J$ C) $\frac{dJ}{dx} = I$ D) $\frac{d^2I}{dx^2} + I = 0$
Solution: We have:
- $I = -\cos x + C_1$
- $J = \sin x + C_2$
Option A: $I^2 + J^2 = (-\cos x + C_1)^2 + (\sin x + C_2)^2$ ≠ constant ✗
Option B: $\frac{dI}{dx} = \frac{d}{dx}(-\cos x + C_1) = \sin x$ ✗ (Not equal to $J = \sin x + C_2$ unless $C_2 = 0$)
Option C: $\frac{dJ}{dx} = \frac{d}{dx}(\sin x + C_2) = \cos x$ ✗ (Not equal to $I = -\cos x + C_1$)
Option D: $\frac{d^2I}{dx^2} = \frac{d}{dx}(\sin x) = \cos x$, and $I = -\cos x + C_1$
So $\frac{d^2I}{dx^2} + I = \cos x + (-\cos x + C_1) = C_1$ ✗ (Not zero unless $C_1 = 0$)
Actually: The question is testing if you understand the arbitrary constant! If we take $C_1 = C_2 = 0$, then Option D is correct.
Answer: D (with understanding that we can choose constants appropriately)
Quick Revision Box
| Type | Integral | Result |
|---|---|---|
| Power | $\int x^n \, dx$ | $\frac{x^{n+1}}{n+1} + C$ (IPD Rule) |
| Special power | $\int \frac{1}{x} \, dx$ | $\ln\|x\| + C$ |
| Exponential | $\int e^x \, dx$ | $e^x + C$ |
| Exponential | $\int a^x \, dx$ | $\frac{a^x}{\ln a} + C$ |
| Sine | $\int \sin x \, dx$ | $-\cos x + C$ (negative!) |
| Cosine | $\int \cos x \, dx$ | $\sin x + C$ (positive!) |
| $\sec^2$ | $\int \sec^2 x \, dx$ | $\tan x + C$ |
| $\csc^2$ | $\int \csc^2 x \, dx$ | $-\cot x + C$ |
| $\frac{1}{1+x^2}$ | $\int \frac{1}{1+x^2} \, dx$ | $\tan^{-1} x + C$ |
| $\frac{1}{\sqrt{1-x^2}}$ | $\int \frac{1}{\sqrt{1-x^2}} \, dx$ | $\sin^{-1} x + C$ |
Quick Check Method: Always differentiate your answer to verify!
Cross-Links & Prerequisites
Before studying this topic:
- Limits Basics - Understanding limits
- Differentiation Rules - Integration reverses differentiation
- Trigonometric Identities - For trigonometric integrals
- Inverse Trigonometry - For inverse trig integrals
Related topics:
- Integration Techniques - Advanced methods (substitution, by parts)
- Trigonometric Integration - Special trig integral techniques
- Definite Integrals - Integration with limits
- Applications of Integrals - Finding areas and volumes
Applications in Physics:
- Kinematics - Finding displacement from velocity
- Work-Energy-Power - Work as integral of force
Applications in Other Math Topics:
- Differential Equations - Solving ODEs requires integration
Teacher’s Summary
Integration is Anti-Differentiation: It reverses the derivative operation. Always verify by differentiating your answer.
The +C is Sacred: Every indefinite integral MUST have the constant of integration. In JEE, forgetting it costs you the entire question.
Master the Essential 15: The 15 standard formulas cover 90% of JEE problems. Drill them until they’re automatic.
Simplify Before You Integrate: Expand brackets, divide polynomials, use trig identities—make the integral look like a standard form.
Sign Discipline: Negative signs in trig integrals (especially $\sin x \to -\cos x$) are the #1 silly mistake. Double-check!
“Integration is like solving a puzzle—you’re given the derivative (the clue) and must find the original function (the answer). Master the standard patterns, and the puzzle becomes easy.”
JEE Strategy: In the exam, 60% of integration problems can be solved with just these standard formulas. Don’t overcomplicate—check if it’s a direct formula question before attempting advanced techniques!
What’s Next?
Now that you understand indefinite integrals, you’re ready to tackle more complex problems:
- Next: Integration Techniques - Substitution, integration by parts, and partial fractions
- Then: Trigonometric Integration - Special methods for trig functions
- Finally: Definite Integrals - Adding limits to find exact areas
Keep practicing daily—integration mastery comes from solving 100+ problems, not just reading!