📘 Calculus · Chapter MN14🎯 NDA Level : High Priority
Integration is the reverse of differentiation — it accumulates quantities and computes areas under curves. For NDA, integral calculus covers two major parts: indefinite integrals (finding the antiderivative, expressed with +C) and definite integrals (computing a precise numerical value using limits). Both parts are tested every year with a mix of direct formula application and technique-based problems.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Direct integration using standard formulae (xn, sin x, ex, 1/x, etc.); (2) Integration by substitution: recognising f(g(x))⋅g′(x) patterns; (3) Integration by parts: products like x ex, x sin x, x ln x; (4) Partial fractions: rational functions with factorable denominators; (5) Evaluating definite integrals using the Fundamental Theorem of Calculus; (6) Applying properties of definite integrals, especially the king property; (7) Using even/odd function properties to simplify ∫−aa f(x)dx; (8) Area under curves as a simple definite integral application.
Topics at a Glance
① Antiderivatives & Standard Integrals
xn, sin x, ex, 1/x, complete table
② Integration by Substitution
Let u = g(x), du = g′(x) dx
③ Integration by Parts
∫u dv = uv − ∫v du (ILATE rule)
④ Partial Fractions
Decompose rational functions
⑤ Definite Integrals & FTC
∫ab f(x) dx = F(b) − F(a)
⑥ Properties of Definite Integrals
King property, even/odd, split interval
1. Antiderivatives & Standard Integrals
1.1
Integration as the Reverse of Differentiation
The indefinite integral gives a family of functions differing by a constant C
If d/dx[F(x)] = f(x), then ∫f(x)dx = F(x) + C, where C is the constant of integration. Integration undoes differentiation — every derivative rule has a corresponding integral rule.
∫ f(x) dx = F(x) + C
F(x) is any antiderivative of f(x). C is the constant of integration (arbitrary).
The +C is mandatory in indefinite integrals — it represents an entire family of parallel curves.
⚡ Complete Standard Integrals Table
ALGEBRAIC:
∫ xⁿ dx = x^(n+1)/(n+1) + C (n ≠ −1)
∫ dx = x + C
∫ k dx = kx + C
∫ 1/x dx = ln|x| + C (note: absolute value!)
∫ 1/x² dx = −1/x + C
∫ √x dx = (2/3)x^(3/2) + C
TRIGONOMETRIC:
∫ sin x dx = −cos x + C
∫ cos x dx = sin x + C
∫ sec²x dx = tan x + C
∫ cosec²x dx = −cot x + C
∫ sec x tan x dx = sec x + C
∫ cosec x cot x dx = −cosec x + C
EXPONENTIAL & LOGARITHMIC:
∫ eˣ dx = eˣ + C
∫ aˣ dx = aˣ / ln a + C (a > 0, a ≠ 1)
∫ ln x dx = x ln x − x + C (integration by parts result)
INVERSE TRIGONOMETRIC:
∫ 1/√(1−x²) dx = sin⁻¹ x + C
∫ 1/(1+x²) dx = tan⁻¹ x + C
∫ 1/(√(a²−x²)) dx = sin⁻¹(x/a) + C
∫ 1/(a²+x²) dx = (1/a) tan⁻¹(x/a) + C
Power rule memory: bring up the exponent by 1, divide by the new exponent. The ONE exception is n = −1: ∫ x−1 dx = ln|x| + C (not 0 in denominator).
Signs to Watch
∫ sin x dx = −cos x + C (negative!)
∫ cos x dx = +sin x + C (positive)
∫ cosec² x dx = −cot x + C (negative!)
∫ cosec x cot x dx = −cosec x + C
Signs are opposite to their derivatives
Generalised Forms (Chain Rule Reverse)
∫ sin(ax+b) dx = −cos(ax+b)/a + C
∫ cos(ax+b) dx = sin(ax+b)/a + C
∫ eax+b dx = eax+b/a + C
∫ (ax+b)ⁿ dx = (ax+b)^(n+1)/(a(n+1)) + C
∫ 1/(ax+b) dx = ln|ax+b|/a + C
📝 TOPIC-WISE PYQ
Standard Integrals — NDA-Pattern Questions
Q1. ∫(x³ + 2x − 5) dx = ?
(a) x⁴/4 + x² − 5x + C (b) 3x²+2 (c) x⁴/4 + 2x − 5 + C (d) x⁴ + x² − 5x + C
(a) e3x + C (b) 3e3x + C (c) e3x/3 + C (d) e3x−1 + C
Answer: (c) e3x/3 + C
∫ eax dx = eax/a + C. Here a = 3: e3x/3 + C.
Q3. ∫ cos(2x + 1) dx = ?
(a) sin(2x+1) + C (b) −sin(2x+1)/2 + C (c) sin(2x+1)/2 + C (d) 2sin(2x+1) + C
Answer: (c) sin(2x+1)/2 + C
∫ cos(ax+b) dx = sin(ax+b)/a + C. Here a=2: sin(2x+1)/2 + C.
2. Techniques of Integration
2.1
Integration by Substitution
Replace a complicated expression with a simple variable u
Substitution is the integration equivalent of the chain rule. When you see f(g(x)) ⋅ g′(x), let u = g(x) so du = g′(x) dx and the integral simplifies.
⚡ Substitution Method — Steps & Recognition
Step 1: Choose u = g(x) (usually the “inside” function).
Step 2: Find du/dx = g′(x), so du = g′(x) dx.
Step 3: Substitute: replace g(x) with u and g′(x)dx with du.
Step 4: Integrate in terms of u (use standard formula).
Step 5: Back-substitute: replace u with g(x).
Recognition patterns:
∫ f(ax+b) dx → let u = ax+b
∫ f(xⁿ) ⋅ x^(n-1) dx → let u = xⁿ
∫ f(sin x) cos x dx → let u = sin x
∫ f(cos x) sin x dx → let u = cos x
∫ f(ln x) / x dx → let u = ln x
∫ f(eˣ) eˣ dx → let u = eˣ
∫ f(x) / g(x) dx where numerator is derivative of denominator:
= ln|g(x)| + C (very common NDA type)
The key recognition: if the numerator is proportional to the derivative of the denominator, the answer is ln|denominator| + C. This covers all ∫ f′/f forms.
Worked Example — Substitution
Evaluate ∫ 2x sin(x²) dx.
Let u = x². Then du = 2x dx. The integral becomes:
∫ sin(u) du = −cos(u) + C = −cos(x²) + C.
Worked Example — Logarithmic Form
Evaluate ∫ (2x+3)/(x²+3x+5) dx.
Numerator = d/dx(x²+3x+5). So this is ∫ f′(x)/f(x) dx = ln|f(x)| + C.
Result: ln|x²+3x+5| + C.
2.2
Integration by Parts — ILATE Rule
For products of two functions — ILATE tells you which to differentiate
Integration by parts is the integration equivalent of the product rule. It reduces a product integral into a simpler one.
⚡ Integration by Parts Formula & ILATE Rule
FORMULA:
∫ u ⋅ v dx = u ⋅ (∫v dx) − ∫[du/dx ⋅ ∫v dx] dx
In short: ∫ u dv = uv − ∫ v du
ILATE Rule (priority order for choosing u):
I — Inverse trigonometric functions (sin⁻¹x, tan⁻¹x, etc.)
L — Logarithmic functions (ln x, log x)
A — Algebraic functions (x, x², polynomials)
T — Trigonometric functions (sin x, cos x)
E — Exponential functions (eˣ, aˣ)
The function appearing EARLIER in ILATE is chosen as u.
The other function becomes dv (and is integrated to get v).
Special case — ∫ eˣ [f(x) + f′(x)] dx:
= eˣ f(x) + C (direct result, no calculation needed)
ILATE tells you which part to differentiate (u) and which to integrate (v). The exponential ex[f(x)+f′(x)] form is the most-tested by-parts result in NDA — recognise and apply directly.
Worked Example — Integration by Parts
Evaluate ∫ x ex dx.
ILATE: A (x) comes before E (ex) → u = x, dv = ex dx.
du = dx, v = ex.
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C = ex(x − 1) + C.
Worked Example — Integration by Parts (ln x)
Evaluate ∫ ln x dx.
Write as ∫ ln x ⋅ 1 dx. ILATE: L (ln x) before A (1) → u = ln x, dv = dx.
du = (1/x) dx, v = x.
∫ ln x dx = x ln x − ∫ x ⋅ (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.
Worked Example — Direct ex[f+f′] Form
Evaluate ∫ ex(sin x + cos x) dx.
Recognise: f(x) = sin x, f′(x) = cos x. This is exactly ex[f(x)+f′(x)].
Direct result: ex sin x + C.
2.3
Integration by Partial Fractions
Break a complicated rational function into simpler fractions that can be integrated directly
⚡ Partial Fractions — Standard Forms
CASE 1 — Non-repeated linear factors in denominator:
P(x)/[(x−a)(x−b)] = A/(x−a) + B/(x−b)
→ multiply both sides by denominator, compare coefficients or substitute x=a, x=b.
CASE 2 — Repeated linear factor:
P(x)/(x−a)² = A/(x−a) + B/(x−a)²
CASE 3 — Irreducible quadratic in denominator:
P(x)/[(x−a)(x²+bx+c)] = A/(x−a) + (Bx+C)/(x²+bx+c)
(numerator of quadratic factor must have degree one less than denominator)
After partial fraction decomposition:
∫ A/(x−a) dx = A ln|x−a| + C
∫ B/(x−a)² dx = −B/(x−a) + C
∫ (Bx+C)/(x²+k²) dx → split into ln and tan⁻¹ parts
Condition: Degree of P(x) must be LESS than degree of denominator.
If not → first do polynomial long division, then apply partial fractions.
For NDA, only Case 1 (two distinct linear factors) is regularly tested. Case 2 and Case 3 appear occasionally. Always factorise the denominator completely before setting up partial fractions.
Worked Example — Partial Fractions
Evaluate ∫ 1/[(x−1)(x+2)] dx.
Decompose: 1/[(x−1)(x+2)] = A/(x−1) + B/(x+2).
Multiply: 1 = A(x+2) + B(x−1).
x=1: 1 = 3A → A = 1/3. x=−2: 1 = −3B → B = −1/3.
∫ = (1/3)∫1/(x−1) dx − (1/3)∫1/(x+2) dx
= (1/3) ln|x−1| − (1/3) ln|x+2| + C = (1/3) ln|(x−1)/(x+2)| + C.
📝 TOPIC-WISE PYQ
Integration Techniques — NDA-Pattern Questions
Q4. ∫ x/(x²+1) dx = ?
(a) ln(x²+1) + C (b) (1/2)ln(x²+1) + C (c) 2ln(x²+1) + C (d) tan⁻¹x + C
Answer: (b) (1/2)ln(x²+1) + C
Numerator x = (1/2)(2x) = (1/2) ⋅ d/dx(x²+1).
∫ f′/f = ln|f|: ∫ x/(x²+1) dx = (1/2)ln(x²+1) + C.
Q5. ∫ x cos x dx = ?
(a) x sin x + cos x + C (b) x sin x − cos x + C (c) cos x − x sin x + C (d) sin x + C
Answer: (a) x sin x + cos x + C
By parts: u=x, dv=cos x dx. du=dx, v=sin x.
∫ x cos x dx = x sin x − ∫ sin x dx = x sin x −(−cos x) = x sin x + cos x + C.
Q6. ∫ ex(1/x + ln x) dx = ?
(a) ex ln x + C (b) ex/x + C (c) xex ln x + C (d) ex(ln x − 1) + C
Answer: (a) ex ln x + C
Form: ex[f(x)+f′(x)] where f(x) = ln x and f′(x) = 1/x.
Direct result: ex ln x + C.
Q7. ∫ sin² x dx = ?
(a) x/2 − sin(2x)/4 + C (b) x/2 + sin(2x)/4 + C (c) −cos(2x)/2 + C (d) sin(2x)/2 + C
Answer: (a) x/2 − sin(2x)/4 + C
Use identity: sin² x = (1 − cos 2x)/2.
∫ (1−cos 2x)/2 dx = x/2 − sin(2x)/4 + C = x/2 − sin(2x)/4 + C.
🔥 TRICKY QUESTIONS
Integration Techniques — Classic NDA Traps
🤯 T1. Evaluate ∫ 1/(sin x cos x) dx.
Multiply and divide by 2: 1/(sin x cos x) = 2/(2 sin x cos x) = 2/sin(2x) = 2 cosec(2x).
∫ 2 cosec(2x) dx = 2 ⋅ (1/2) ln|tan(x)| + C = ln|tan x| + C.
Alternatively: divide numerator and denominator by cos²x: ∫ sec²x/tan x dx = ∫ d(tan x)/tan x = ln|tan x| + C. Strategy: When the integrand has mixed trig, try dividing by cosⁿx or use the 2sin x cos x = sin 2x identity.
🤯 T2. Evaluate ∫ (x+1)/(x²+2x+3) dx.
Observe: d/dx(x²+2x+3) = 2x+2 = 2(x+1).
So (x+1) = (1/2)(2x+2) = (1/2) d/dx(x²+2x+3).
∫ (x+1)/(x²+2x+3) dx = (1/2) ∫ (2x+2)/(x²+2x+3) dx = (1/2) ln|x²+2x+3| + C. Tip: When numerator is proportional to derivative of denominator, use the f′/f = ln|f| shortcut. Always check this first before trying any other technique.
🤯 T3. Evaluate ∫ ex(2+sin 2x)/(1+cos 2x) dx.
Use identities: 1+cos 2x = 2cos²x and sin 2x = 2 sin x cos x.
= ∫ ex(2 + 2 sin x cos x) / (2 cos²x) dx
= ∫ ex[1/cos²x + sin x/cos x] dx
= ∫ ex[sec²x + tan x] dx.
Recognise: f(x) = tan x, f′(x) = sec²x → form ex[f′(x)+f(x)].
Result: ex tan x + C. Key insight: Always simplify the integrand using trig identities BEFORE choosing a technique.
3. Definite Integrals & Fundamental Theorem of Calculus
3.1
FTC & Geometric Meaning — Area Under a Curve
F(b) − F(a): substitute the limits into the antiderivative
The definite integral of f(x) from a to b gives the net signed area between the curve y = f(x) and the x-axis over [a, b]. It is computed using the Fundamental Theorem of Calculus (FTC).
⚡ Fundamental Theorem of Calculus
If F(x) is an antiderivative of f(x) [i.e., F′(x) = f(x)], then:
∫ab f(x) dx = [F(x)]ₘ=a to b = F(b) − F(a)
Steps:
1. Find the indefinite integral F(x) (omit the +C for definite integrals).
2. Evaluate F(b) − F(a).
3. The result is a NUMBER (no variable, no +C).
Geometric meaning:
• If f(x) ≥ 0 on [a,b]: the value = area under the curve (positive).
• If f(x) ≤ 0 on [a,b]: the value = negative of area (negative).
• Net area = algebraic sum; total area = ∫|f(x)|dx (always positive).
Area formula:
Area between y=f(x) and x-axis on [a,b] = ∫ab |f(x)| dx
The constant of integration always cancels out in definite integrals: [F(x)+C] evaluated from a to b = F(b)+C − F(a)−C = F(b)−F(a). So +C is never needed.
Fig 1: The definite integral ∫ab f(x) dx equals the shaded area between the curve and the x-axis (when f(x) ≥ 0).
Worked Example — Evaluating a Definite Integral
Evaluate ∫02 (3x² + 2x − 1) dx.
F(x) = x³ + x² − x.
F(2) = 8 + 4 − 2 = 10. F(0) = 0.
∫02 = F(2) − F(0) = 10 − 0 = 10.
4. Properties of Definite Integrals
4.1
All Properties — With Worked Applications
The King property and even/odd property are the most tested in NDA
P7 — Periodic function (period T):
∫0nT f(x) dx = n ∫0T f(x) dx
P8 — Half-interval symmetry:
∫02a f(x) dx = 2∫0a f(x) dx if f(2a−x) = f(x) [symmetric about x=a]
∫02a f(x) dx = 0 if f(2a−x) = −f(x) [anti-symmetric]
King property for [0, π/2]:
∫0π/2 f(sin x) dx = ∫0π/2 f(cos x) dx
[Replace x with π/2 − x; sin ↔ cos]
Special result:
∫0π/2 sinⁿx dx = ∫0π/2 cosⁿx dx (same value for any n)
The King property is the most powerful tool for NDA definite integral problems. When the integral looks hard, apply King (replace x → a+b−x), add to the original, and both sides often simplify dramatically.
Worked Example — King Property
Evaluate I = ∫0π x sin x / (1 + cos²x) dx.
Apply King: replace x → π−x. sin(π−x) = sin x, cos(π−x) = −cos x, cos²(π−x) = cos²x.
I = ∫0π (π−x) sin x / (1+cos²x) dx.
Add both: 2I = ∫0π π sin x / (1+cos²x) dx = π ∫0π sin x/(1+cos²x) dx.
Let u = cos x: du = −sin x dx. Limits: x=0 → u=1; x=π → u=−1.
Definite Integrals — King Property & Symmetry Traps
🤯 T4. Evaluate ∫0π ln(sin x) dx.
Let I = ∫0π ln(sin x) dx.
King (replace x → π−x): sin(π−x) = sin x, so the integrand is unchanged.
Thus I = ∫0π ln(sin(π−x)) dx = I. (This confirms symmetry but doesn’t solve it.)
Use the half-interval: I = 2∫0π/2 ln(sin x) dx.
The known result: ∫0π/2 ln(sin x) dx = −π ln 2 / 2.
So I = 2(−π ln 2 / 2) = −π ln 2. This is a classic NDA result that should be memorised as a standard result.
🤯 T5. If ∫0k x dx = 2∫02 x dx, find k.
∫0k x dx = [x²/2]0k = k²/2.
∫02 x dx = [x²/2]02 = 2.
Equation: k²/2 = 2(2) = 4 → k² = 8 → k = 2√2 (taking positive value). k = 2√2. Always evaluate each side independently, then solve the resulting algebraic equation for k.
🤯 T6. Evaluate ∫0π/2 sin⁶x / (sin⁶x + cos⁶x) dx.
Let I = ∫0π/2 sin⁶x / (sin⁶x + cos⁶x) dx.
King (replace x → π/2 − x): sin x ↔ cos x.
I = ∫0π/2 cos⁶x / (cos⁶x + sin⁶x) dx.
Add both: 2I = ∫0π/2 [sin⁶x + cos⁶x] / [sin⁶x + cos⁶x] dx = ∫0π/2 1 dx = π/2. I = π/4. This works for ANY exponent n — the result is always π/4. A very common NDA question type.
📝 Master Formula Sheet — MN14 Integral Calculus
All critical formulae for rapid pre-exam revision.
∫ Standard Integrals (Algebraic & Exp)
∫xⁿ dx = x^(n+1)/(n+1) + C (n ≠ −1)
∫1/x dx = ln|x| + C
∫eˣ dx = eˣ + C
∫aˣ dx = aˣ/ln a + C
∫(ax+b)ⁿ dx = (ax+b)^(n+1)/[a(n+1)] + C
∫ Standard Integrals (Trig)
∫sin x dx = −cos x + C
∫cos x dx = sin x + C
∫sec²x dx = tan x + C
∫cosec²x dx = −cot x + C
∫1/√(1−x²) dx = sin⁻¹x + C
∫1/(1+x²) dx = tan⁻¹x + C
∬ Techniques
Substitution: let u=g(x), du=g′dx
f′/f form: ∫f′(x)/f(x)dx = ln|f(x)|+C
By parts: ∫u dv = uv − ∫v du (ILATE)
eˣ[f+f′] = eˣf(x) + C (direct)
Partial fractions: decompose, then integrate
♫ Definite Integral Properties
FTC: ∫abf dx = F(b)−F(a)
Reversal: ∫abf = −∫baf
King: ∫abf(x)dx = ∫abf(a+b−x)dx
Odd on [−a,a]: integral = 0
Even on [−a,a]: integral = 2∫0a
♔ King Property Applications
∫0π/2sin x/(sin+cos) dx = π/4
∫0π/2sinⁿx/(sinⁿ+cosⁿ) dx = π/4
∫0π/2sin x dx = ∫0π/2cos x dx = 1
∫0πx f(sin x) dx = (π/2)∫0πf(sin x)dx
🌟 Key Results to Memorise
∫sin²x dx = x/2 − sin(2x)/4 + C
∫cos²x dx = x/2 + sin(2x)/4 + C
∫ln x dx = x ln x − x + C
∫0π/2ln(sin x)dx = −π ln2/2
∫aaf dx = 0 always
⚡ Quick Revision Booster — MN14 Integral Calculus
∫ Standard Quick-Ref
∫xⁿ → x^(n+1)/(n+1) (n≠−1)
∫1/x = ln|x| (need absolute value)
∫sin = −cos; ∫cos = +sin
∫eˣ = eˣ; ∫aˣ = aˣ/ln a
∫ on [a,a] = 0 always
🔗 Substitution
Numerator = derivative of denominator → ln|denom| + C
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