Module 5 · JEE Mathematics
Integration — The Reverse of Everything
Undoing derivatives, adding up tiny slices, finding areas — integration is where calculus really shines.
🔄 Indefinite
📏 Definite
📐 Area
⚡ Techniques
🎯 Practice
🔄 Part 1: Indefinite Integration — The Anti-Derivative
📏 Part 2: Definite Integration — Adding Up the Slices
📐 Part 3: Area Under Curves — Integration in Action
⚡ Part 4: Advanced Techniques — When Basic Methods Aren't Enough
⚠️ Common Mistakes (Don't Fall For These!)
🧠 Integration Toolkit
🔥 Advanced Worked Examples (JEE Level)
🎯 Practice Zone: Test Your Understanding
Integration Asks: "What Function Did This Come From?"
If differentiation tells you the rate of change, integration tells you the original quantity. It's like knowing someone's speed at every moment and asking: where did they start?
🔄 The Core Idea
Integration = Reverse Differentiation — Like Watching a Movie Backward
If f(x) = x², then f'(x) = 2x. Integration goes the other way: ∫ 2x dx = x² + C.
The integral sign ∫ means "find the anti-derivative." The dx tells you the variable. The + C is the constant of integration — because the derivative of a constant is zero, we can never know the exact constant from just the derivative.
Analogy: If you know a car's speed at every moment (derivative), you can figure out where it went (integral), but you won't know exactly where it started (the + C).
The integral sign ∫ means "find the anti-derivative." The dx tells you the variable. The + C is the constant of integration — because the derivative of a constant is zero, we can never know the exact constant from just the derivative.
Analogy: If you know a car's speed at every moment (derivative), you can figure out where it went (integral), but you won't know exactly where it started (the + C).
💡 ∫ f'(x) dx = f(x) + C. The integral undoes the derivative, but the + C represents lost information.
The Basic Integrals — The Building Blocks
Every integration problem ultimately breaks down into these basic forms. Learn the PATTERN, not the formula.
📊 Basic Integral Patterns — The Building Blocks
The 10 basic integrals form the foundation of ALL integration. Every complex integral is a combination of these patterns using substitution, parts, or partial fractions.
Worked Example: Basic Integration
Find ∫ (3x² + 2x − 5) dx.
1
Integrate term by term using linearity
▶
∫ 3x² dx = 3·x³/3 = x³
∫ 2x dx = 2·x²/2 = x²
∫ −5 dx = −5x
∫ 2x dx = 2·x²/2 = x²
∫ −5 dx = −5x
∫ (3x² + 2x − 5) dx = x³ + x² − 5x + C
Check by differentiating: d/dx(x³ + x² − 5x + C) = 3x² + 2x − 5 ✅
Integration by Substitution — The Chain Rule in Reverse
The chain rule says: d/dx f(g(x)) = f'(g(x))·g'(x). Substitution reverses this: if you spot a function AND its derivative in the integrand, substitute.
🔄 Substitution
If You See f(g(x))·g'(x), Let u = g(x)
Example: ∫ 2x·cos(x²) dx
Notice: the derivative of x² is 2x, which appears right next to cos(x²). Let u = x², then du/dx = 2x, so du = 2x dx.
The integral becomes: ∫ cos(u) du = sin(u) + C = sin(x²) + C.
The pattern: Look for a function (the "inner" function) whose derivative is also present (up to a constant factor). Substitute u = inner function, replace dx with du/derivative.
Notice: the derivative of x² is 2x, which appears right next to cos(x²). Let u = x², then du/dx = 2x, so du = 2x dx.
The integral becomes: ∫ cos(u) du = sin(u) + C = sin(x²) + C.
The pattern: Look for a function (the "inner" function) whose derivative is also present (up to a constant factor). Substitute u = inner function, replace dx with du/derivative.
💡 Substitution = "spot the chain rule in reverse." If a function and its neighbor look like a chain, let u = the inner function.
Integration by Parts — The Product Rule in Reverse
The product rule: (uv)' = u'v + uv'. Rearranging: ∫ uv' dx = uv − ∫ u'v dx.
⚡ Integration by Parts
∫ u dv = uv − ∫ v du — The Product Rule in Reverse
Choose u as the function that simplifies when differentiated, and dv as the function that doesn't get more complicated when integrated.
ILATE priority for choosing u:
I — Inverse trig (e.g., sin⁻¹(x))
L — Logarithmic (e.g., ln(x))
A — Algebraic (e.g., x, x²)
T — Trigonometric (e.g., sin(x), cos(x))
E — Exponential (e.g., eˣ)
Example: ∫ x·eˣ dx. u = x (algebraic), dv = eˣ dx (exponential).
du = dx, v = eˣ. Formula: uv − ∫ v du = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C.
ILATE priority for choosing u:
I — Inverse trig (e.g., sin⁻¹(x))
L — Logarithmic (e.g., ln(x))
A — Algebraic (e.g., x, x²)
T — Trigonometric (e.g., sin(x), cos(x))
E — Exponential (e.g., eˣ)
Example: ∫ x·eˣ dx. u = x (algebraic), dv = eˣ dx (exponential).
du = dx, v = eˣ. Formula: uv − ∫ v du = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C.
💡 ILATE tells you which function to set as u. The other function becomes dv. Then: ∫ u dv = uv − ∫ v du.
🌟 Indefinite Integration Cheat Sheet
Power rule: ∫ xⁿ = xⁿ⁺¹/(n+1). | Substitution: ∫ f(g(x))g'(x) dx = ∫ f(u) du
Integration by parts: ∫ u dv = uv − ∫ v du | Choose u by ILATE order
The Definite Integral = The Area Under the Curve
While indefinite integration gives a family of functions, definite integration gives a NUMBER — the net area under the curve between two x-values.
📏 The Definite Integral = Area of Infinitely Thin Rectangles
The Riemann sum: The definite integral ∫ₐᵇ f(x) dx is the limit of the sum of rectangle areas as the width → 0. Each rectangle has height f(xᵢ) and width Δx. As Δx → 0, the approximation becomes exact.
The Fundamental Theorem of Calculus (FTC)
The FTC connects differentiation and integration — it says that integration (adding up slices) is the opposite of differentiation (finding slopes).
📏 The FTC
∫ₐᵇ f(x) dx = F(b) − F(a), where F'(x) = f(x)
This is THE most important theorem in calculus. It says:
1. Find the anti-derivative F(x) of f(x).
2. Evaluate F at the upper limit b.
3. Evaluate F at the lower limit a.
4. Subtract: F(b) − F(a).
Example: ∫₀² x² dx = [x³/3]₀² = (8/3) − (0/3) = 8/3.
Properties:
• ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx (reversing limits flips the sign)
• ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx (splitting intervals)
• ∫ₐᵃ f(x) dx = 0 (zero-width interval)
1. Find the anti-derivative F(x) of f(x).
2. Evaluate F at the upper limit b.
3. Evaluate F at the lower limit a.
4. Subtract: F(b) − F(a).
Example: ∫₀² x² dx = [x³/3]₀² = (8/3) − (0/3) = 8/3.
Properties:
• ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx (reversing limits flips the sign)
• ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx (splitting intervals)
• ∫ₐᵃ f(x) dx = 0 (zero-width interval)
💡 FTC = "Find the anti-derivative, plug in the bounds, subtract." That's all definite integration is!
The King's Property — ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx
This property is a JEE favorite because it simplifies integrals that look impossible. The idea: replace x with (a + b − x) — the integral stays the same.
👑 King's Property
∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx — The Reflection Trick
Why it works: As x goes from a to b, (a + b − x) goes from b to a. The integral over the same interval of the reflected function equals the original integral.
Example: ∫₀^{π/2} sin(x)/(sin(x) + cos(x)) dx.
Let I = ∫₀^{π/2} sin(x)/(sin(x) + cos(x)) dx.
Using King's: I = ∫₀^{π/2} sin(π/2 − x)/(sin(π/2 − x) + cos(π/2 − x)) dx
= ∫₀^{π/2} cos(x)/(cos(x) + sin(x)) dx.
Adding: 2I = ∫₀^{π/2} (sin + cos)/(sin + cos) dx = ∫₀^{π/2} 1 dx = π/2.
Therefore: I = π/4. Magic! 🎩
Example: ∫₀^{π/2} sin(x)/(sin(x) + cos(x)) dx.
Let I = ∫₀^{π/2} sin(x)/(sin(x) + cos(x)) dx.
Using King's: I = ∫₀^{π/2} sin(π/2 − x)/(sin(π/2 − x) + cos(π/2 − x)) dx
= ∫₀^{π/2} cos(x)/(cos(x) + sin(x)) dx.
Adding: 2I = ∫₀^{π/2} (sin + cos)/(sin + cos) dx = ∫₀^{π/2} 1 dx = π/2.
Therefore: I = π/4. Magic! 🎩
💡 King's property: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx. In JEE, this turns impossible-looking trig integrals into trivial ones.
Even & Odd Function Integration
For symmetric intervals [−a, a], the parity of the function simplifies things dramatically.
📊 Even & Odd — Half the Work
Even functions: Symmetric about the y-axis → integrate from 0 to a and double it. Odd functions: Symmetric about the origin → the integral over [−a, a] is always zero because the left and right halves cancel.
🌟 Definite Integration: Key Properties
FTC: ∫ₐᵇ f(x) dx = F(b) − F(a) | King's: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx
Even: ∫₋ₐᵃ = 2∫₀ᵃ | Odd: ∫₋ₐᵃ = 0 | Splitting: ∫ₐᵇ + ∫ᵇᶜ = ∫ₐᶜ
The Definite Integral = The Net Area (Above − Below the x-axis)
This is the most visual application of integration: finding the area bounded by curves.
📐 Area Between Curves
Area = ∫ (Top Curve − Bottom Curve) dx
Area between y = f(x) and y = g(x) from x = a to x = b:
Steps:
1. Find the intersection points of the curves (where f(x) = g(x)).
2. Determine which curve is on top in each interval.
3. Integrate (top − bottom) over each interval.
4. Add the absolute values.
Area between curve and x-axis: ∫ₐᵇ |f(x)| dx. If f(x) is always positive on the interval, it's just ∫ₐᵇ f(x) dx.
Area = ∫ₐᵇ |f(x) − g(x)| dx
Where f(x) is the upper curve and g(x) is the lower curve on the interval.Steps:
1. Find the intersection points of the curves (where f(x) = g(x)).
2. Determine which curve is on top in each interval.
3. Integrate (top − bottom) over each interval.
4. Add the absolute values.
Area between curve and x-axis: ∫ₐᵇ |f(x)| dx. If f(x) is always positive on the interval, it's just ∫ₐᵇ f(x) dx.
💡 Area = ∫ (top − bottom). If curves cross, split the interval at the intersection points and handle each piece separately.
Partial Fractions, Special Integrals, and Reduction Formulas
These techniques appear frequently in JEE Advanced. Here's a conceptual overview — the formulas are in your toolkit.
Partial Fractions — Breaking Rational Functions into Simple Pieces
When integrating rational functions (polynomial/polynomial), you decompose them into simpler fractions that you know how to integrate.
⚡ Partial Fractions
1/(x² − 1) = ½·1/(x−1) − ½·1/(x+1)
Example: ∫ 1/(x² − 1) dx.
Step 1: Factor denominator: x² − 1 = (x − 1)(x + 1).
Step 2: Write 1/((x−1)(x+1)) = A/(x−1) + B/(x+1).
Step 3: Solve: A(x+1) + B(x−1) = 1. Put x=1: 2A = 1 → A = ½. Put x=−1: −2B = 1 → B = −½.
Step 4: Integrate: ∫ ½·1/(x−1) − ½·1/(x+1) dx = ½ ln|x−1| − ½ ln|x+1| + C = ½ ln|(x−1)/(x+1)| + C.
Partial fractions turns division into addition — then you just integrate ln terms!
Step 1: Factor denominator: x² − 1 = (x − 1)(x + 1).
Step 2: Write 1/((x−1)(x+1)) = A/(x−1) + B/(x+1).
Step 3: Solve: A(x+1) + B(x−1) = 1. Put x=1: 2A = 1 → A = ½. Put x=−1: −2B = 1 → B = −½.
Step 4: Integrate: ∫ ½·1/(x−1) − ½·1/(x+1) dx = ½ ln|x−1| − ½ ln|x+1| + C = ½ ln|(x−1)/(x+1)| + C.
Partial fractions turns division into addition — then you just integrate ln terms!
💡 Partial fractions = "split a complicated fraction into simple ones you know how to integrate." Always factor the denominator first!
Special Integral Forms
These three forms appear constantly in JEE. Derive them once, then use them as formulas.
📊 Three Special Integral Forms
These three special forms (and their variants) cover most JEE integration problems involving √ and quadratic denominators. The key trick: complete the square in the quadratic to express it as a² ± x² or x² ± a².
🌟 Integration Strategy
1. Basic form? → Power rule / trig / exponential. 2. Chain rule pattern? → Substitution. 3. Product? → By parts (ILATE). 4. Rational? → Partial fractions. 5. √ or (a²+x²)? → Special forms.
Always check the form FIRST, then pick the method. Most JEE problems test your ability to CHOOSE the right technique.
These Traps Trip Up Even Top Students
⚠️ Mistake 1
Forgetting the + C in Indefinite Integration
Wrong: ∫ 2x dx = x² ❌
Right: ∫ 2x dx = x² + C ✅
The + C is NOT optional! Since d/dx(constant) = 0, any constant could have been in the original function. The integral represents a family of functions, not just one.
In definite integration, the + C cancels out: F(b) + C − (F(a) + C) = F(b) − F(a). So you don't need + C in definite integrals, but it's a must for indefinite.
Right: ∫ 2x dx = x² + C ✅
The + C is NOT optional! Since d/dx(constant) = 0, any constant could have been in the original function. The integral represents a family of functions, not just one.
In definite integration, the + C cancels out: F(b) + C − (F(a) + C) = F(b) − F(a). So you don't need + C in definite integrals, but it's a must for indefinite.
⚠️ Never forget the + C in indefinite integrals. Without it, your answer is incomplete — you've only found ONE of infinitely many anti-derivatives!
⚠️ Mistake 2
Confusing ∫ 1/x dx with ln(x) Instead of ln|x|
Wrong: ∫ 1/x dx = ln(x) + C ❌
Right: ∫ 1/x dx = ln|x| + C ✅
The absolute value matters because 1/x is defined for negative x too! ln(x) is only defined for x > 0, but ln|x| works for all x ≠ 0.
Example: ∫₋₂⁻¹ 1/x dx = [ln|x|]₋₂⁻¹ = ln(1) − ln(2) = −ln(2). Without the absolute value, you'd get ln(−1) which doesn't exist!
Right: ∫ 1/x dx = ln|x| + C ✅
The absolute value matters because 1/x is defined for negative x too! ln(x) is only defined for x > 0, but ln|x| works for all x ≠ 0.
Example: ∫₋₂⁻¹ 1/x dx = [ln|x|]₋₂⁻¹ = ln(1) − ln(2) = −ln(2). Without the absolute value, you'd get ln(−1) which doesn't exist!
⚠️ ∫ 1/x dx = ln|x| + C, not ln(x) + C. The absolute value handles negative x correctly.
⚠️ Mistake 3
Forgetting to Change Limits When Substituting in Definite Integrals
Wrong: ∫₀² 2x·cos(x²) dx. Let u = x². Then ∫ cos(u) du = sin(u). Then evaluate sin(u) from 0 to 2. ❌
Right: When x = 0 → u = 0. When x = 2 → u = 4. So the integral is ∫₀⁴ cos(u) du = [sin(u)]₀⁴ = sin(4). ✅
When you substitute in a definite integral, you MUST change the limits to the new variable. Alternatively, substitute back to x at the end and use the original limits.
Right: When x = 0 → u = 0. When x = 2 → u = 4. So the integral is ∫₀⁴ cos(u) du = [sin(u)]₀⁴ = sin(4). ✅
When you substitute in a definite integral, you MUST change the limits to the new variable. Alternatively, substitute back to x at the end and use the original limits.
⚠️ Substitution in definite integrals: change the limits TO the new variable. Forgetting this is the most common definite integration error!
4 Strategies That Crack Almost Any Problem
🧠 Strategy 1
"Identify the Form First" — The Integration Flowchart
Before doing ANY algebra, look at the integrand and ask:
• Is it a basic form (power, trig, exponential)? → Use the basic rules.
• Does it look like f(g(x))·g'(x)? → Substitute u = g(x).
• Is it a product of two different kinds of functions? → Integration by parts (ILATE).
• Is it a rational function (polynomial/polynomial)? → Partial fractions.
• Does it have √(a²−x²) or 1/(a²+x²)? → Special forms.
Most JEE problems test your ability to recognize which technique to use. Practice identifying the form at a glance.
• Is it a basic form (power, trig, exponential)? → Use the basic rules.
• Does it look like f(g(x))·g'(x)? → Substitute u = g(x).
• Is it a product of two different kinds of functions? → Integration by parts (ILATE).
• Is it a rational function (polynomial/polynomial)? → Partial fractions.
• Does it have √(a²−x²) or 1/(a²+x²)? → Special forms.
Most JEE problems test your ability to recognize which technique to use. Practice identifying the form at a glance.
💡 Don't just start computing. First: what kind of integral is this? The form determines the method.
🧠 Strategy 2
"Properties First" for Definite Integrals
Before evaluating a definite integral, check if properties can simplify it:
• Is the interval symmetric [−a, a]? → Check if the function is even or odd.
• Does the integrand look like it could use King's property (especially with trig functions and limits 0 to π/2)?
• Can you split the integral at a point where the function changes form (like |x|)?
Pro tip: If the integral looks impossibly hard, try King's property. If it still looks hard, add the original integral to the King's-transformed version — the sum often simplifies dramatically.
• Is the interval symmetric [−a, a]? → Check if the function is even or odd.
• Does the integrand look like it could use King's property (especially with trig functions and limits 0 to π/2)?
• Can you split the integral at a point where the function changes form (like |x|)?
Pro tip: If the integral looks impossibly hard, try King's property. If it still looks hard, add the original integral to the King's-transformed version — the sum often simplifies dramatically.
💡 Properties of definite integrals are NOT optional shortcuts — they're the main method for JEE. King's property alone solves 30% of JEE definite integral problems.
🧠 Strategy 3
"Complete the Square" for Quadratic Denominators
When integrating rational functions with quadratic denominators that don't factor nicely:
Complete the square to turn it into one of the special forms.
Example: ∫ 1/(x² + 2x + 5) dx.
Complete the square: x² + 2x + 5 = (x + 1)² + 4.
Now: ∫ 1/((x+1)² + 2²) dx = (1/2)·tan⁻¹((x+1)/2) + C.
This technique works for ANY quadratic — just rewrite x² + bx + c as (x + b/2)² + (c − b²/4).
Complete the square to turn it into one of the special forms.
Example: ∫ 1/(x² + 2x + 5) dx.
Complete the square: x² + 2x + 5 = (x + 1)² + 4.
Now: ∫ 1/((x+1)² + 2²) dx = (1/2)·tan⁻¹((x+1)/2) + C.
This technique works for ANY quadratic — just rewrite x² + bx + c as (x + b/2)² + (c − b²/4).
💡 Completing the square turns ANY quadratic into (something)² + (constant)². Then it's just the tan⁻¹ formula.
🧠 Strategy 4
"Check the Derivative" After Integrating
The fastest way to verify an integration is to differentiate your answer. If you get back the original integrand, you're correct.
This catches:
• Sign errors (e.g., ∫ sin(x) dx = −cos(x) + C, not cos(x) + C)
• Coefficient errors (e.g., was it ∫ 2x dx = x² + C or 2x² + C?)
• ln vs ln||absolute value issues
For definite integrals, after finding the anti-derivative, quickly verify by differentiating. Then plug in the bounds.
This catches:
• Sign errors (e.g., ∫ sin(x) dx = −cos(x) + C, not cos(x) + C)
• Coefficient errors (e.g., was it ∫ 2x dx = x² + C or 2x² + C?)
• ln vs ln||absolute value issues
For definite integrals, after finding the anti-derivative, quickly verify by differentiating. Then plug in the bounds.
💡 Differentiation is the best integration checker. If your answer differentiates to the original function, it's correct. Period.
Multi-Step Problems That Combine Multiple Concepts
Example 1: King's Property + Substitution
Evaluate I = ∫₀^{π/2} √(tan(x)) dx.
1
Apply King's property
▶
I = ∫₀^{π/2} √(tan(x)) dx.
By King's property (replace x with π/2 − x):
I = ∫₀^{π/2} √(tan(π/2 − x)) dx = ∫₀^{π/2} √(cot(x)) dx.
Add: 2I = ∫₀^{π/2} [√(tan(x)) + √(cot(x))] dx.
By King's property (replace x with π/2 − x):
I = ∫₀^{π/2} √(tan(π/2 − x)) dx = ∫₀^{π/2} √(cot(x)) dx.
Add: 2I = ∫₀^{π/2} [√(tan(x)) + √(cot(x))] dx.
2
Simplify the integrand
▶
√(tan(x)) + √(cot(x)) = √(tan(x)) + 1/√(tan(x))
= (tan(x) + 1)/√(tan(x)) = (sin(x)/cos(x) + 1)/√(tan(x))
= (sin(x) + cos(x))/√(sin(x)cos(x)).
So 2I = ∫₀^{π/2} (sin(x) + cos(x))/√(sin(x)cos(x)) dx.
= (tan(x) + 1)/√(tan(x)) = (sin(x)/cos(x) + 1)/√(tan(x))
= (sin(x) + cos(x))/√(sin(x)cos(x)).
So 2I = ∫₀^{π/2} (sin(x) + cos(x))/√(sin(x)cos(x)) dx.
3
Substitute t = sin(x) − cos(x)
▶
Let t = sin(x) − cos(x). Then dt = (cos(x) + sin(x)) dx.
t² = sin²(x) + cos²(x) − 2sin(x)cos(x) = 1 − 2sin(x)cos(x).
So sin(x)cos(x) = (1 − t²)/2.
When x = 0: t = 0 − 1 = −1. When x = π/2: t = 1 − 0 = 1.
2I = ∫₋₁¹ dt / √((1 − t²)/2) = √2 ∫₋₁¹ dt/√(1 − t²)
= √2 [sin⁻¹(t)]₋₁¹ = √2(π/2 − (−π/2)) = √2·π.
t² = sin²(x) + cos²(x) − 2sin(x)cos(x) = 1 − 2sin(x)cos(x).
So sin(x)cos(x) = (1 − t²)/2.
When x = 0: t = 0 − 1 = −1. When x = π/2: t = 1 − 0 = 1.
2I = ∫₋₁¹ dt / √((1 − t²)/2) = √2 ∫₋₁¹ dt/√(1 − t²)
= √2 [sin⁻¹(t)]₋₁¹ = √2(π/2 − (−π/2)) = √2·π.
I = π/√2
Example 2: Area Between Curves
Find the area bounded by y = x² and y = √x.
1
Find intersection points
▶
Set x² = √x → x⁴ = x → x⁴ − x = 0 → x(x³ − 1) = 0.
Solutions: x = 0, x = 1.
Intersection points: (0, 0) and (1, 1).
At x = ½: x² = 0.25, √x = 0.707. So √x is above x² on (0, 1).
Solutions: x = 0, x = 1.
Intersection points: (0, 0) and (1, 1).
At x = ½: x² = 0.25, √x = 0.707. So √x is above x² on (0, 1).
2
Integrate (top − bottom)
▶
Area = ∫₀¹ (√x − x²) dx
= ∫₀¹ (x¹/² − x²) dx
= [2x³/²/3 − x³/3]₀¹
= (2/3 − 1/3) − (0 − 0)
= 1/3 square units.
= ∫₀¹ (x¹/² − x²) dx
= [2x³/²/3 − x³/3]₀¹
= (2/3 − 1/3) − (0 − 0)
= 1/3 square units.
Area = 1/3 (approximately 0.333 square units)
The region between y = x² and y = √x from x = 0 to x = 1 has area 1/3.
Example 3: Integration by Parts (JEE Advanced Style)
Evaluate ∫₀^π eˣ·cos(x) dx.
1
Apply integration by parts
▶
Let u = cos(x), dv = eˣ dx.
du = −sin(x) dx, v = eˣ.
I = [eˣ·cos(x)]₀^π − ∫₀^π eˣ·(−sin(x)) dx
= [eˣ·cos(x)]₀^π + ∫₀^π eˣ·sin(x) dx
= (e^π·cos(π) − e⁰·cos(0)) + ∫₀^π eˣ·sin(x) dx
= (e^π·(−1) − 1·1) + ∫₀^π eˣ·sin(x) dx
= −e^π − 1 + ∫₀^π eˣ·sin(x) dx.
du = −sin(x) dx, v = eˣ.
I = [eˣ·cos(x)]₀^π − ∫₀^π eˣ·(−sin(x)) dx
= [eˣ·cos(x)]₀^π + ∫₀^π eˣ·sin(x) dx
= (e^π·cos(π) − e⁰·cos(0)) + ∫₀^π eˣ·sin(x) dx
= (e^π·(−1) − 1·1) + ∫₀^π eˣ·sin(x) dx
= −e^π − 1 + ∫₀^π eˣ·sin(x) dx.
2
Apply integration by parts again
▶
Let J = ∫₀^π eˣ·sin(x) dx.
u = sin(x), dv = eˣ dx. du = cos(x) dx, v = eˣ.
J = [eˣ·sin(x)]₀^π − ∫₀^π eˣ·cos(x) dx
= (e^π·0 − e⁰·0) − I = −I.
So I = −e^π − 1 − I
2I = −(e^π + 1)
u = sin(x), dv = eˣ dx. du = cos(x) dx, v = eˣ.
J = [eˣ·sin(x)]₀^π − ∫₀^π eˣ·cos(x) dx
= (e^π·0 − e⁰·0) − I = −I.
So I = −e^π − 1 − I
2I = −(e^π + 1)
I = −(e^π + 1)/2
🌟 Cyclic Integration
When integrating eˣ·cos(x) or eˣ·sin(x), integration by parts creates a CYCLE. Solve for I algebraically: I = expression − I → 2I = expression → I = expression/2.
This technique is called "cyclic integration" or "solving for the integral." It works whenever the second integration by parts brings back the original integral.
Practice Zone: Think, Don't Compute
These questions test if you truly understand integration concepts.