Integration is the reverse of differentiation. When you differentiate F(x) and get f(x), then integrating f(x) gives you back F(x) — plus a constant. That constant is the entire story of why indefinite integration carries a + C at the end.
Think of it this way: differentiation is like a machine that tells you the slope of a curve at every point. Integration is the same machine running backwards — given the slope everywhere, reconstruct the original curve. But here's the catch: infinitely many curves can have the same slope function, each one shifted vertically. The + C captures all of them at once.
More formally, if d/dx [F(x)] = f(x), then ∫ f(x) dx = F(x) + C. We call F(x) the antiderivative (प्रतिअवकल) of f(x).
The symbol ∫ is an elongated S, historically standing for "summa" (sum), because integration ultimately connects to summing infinitely many infinitesimally thin strips — but at the indefinite stage, you don't need to worry about limits. Limits appear in definite integration.
Why NDA cares about this: The NDA Maths paper regularly places 3-5 questions on integration. Most of them test whether you can quickly recognise which technique applies — substitution, a standard formula, or a trigonometric identity. The ones that seem hard are usually just disguised versions of a formula you already know. Your job in the exam hall is the recognition, not the derivation.
These are non-negotiable. Commit them to memory the same way you memorised your times tables.
| Integrand | Result |
|---|---|
| xⁿ (n ≠ -1) | xⁿ⁺¹/(n+1) + C |
| 1/x | ln|x| + C |
| eˣ | eˣ + C |
| sin x | -cos x + C |
| cos x | sin x + C |
| sec²x | tan x + C |
| cosec²x | -cot x + C |
| sec x tan x | sec x + C |
| cosec x cot x | -cosec x + C |
| 1/(x² + a²) | (1/a) tan⁻¹(x/a) + C |
| 1/√(a² - x²) | sin⁻¹(x/a) + C |
The sign errors on the trig ones are where most NDA candidates drop marks. Notice: ∫sin x dx = -cos x + C (negative), but ∫cos x dx = +sin x + C (positive). The cosec² rule gives -cot x + C. Get these wrong in a hurry and you'll pick the mirror-image wrong answer — which is exactly what the option list is designed to tempt you with.
This is the workhorse technique. The idea: if the integrand contains a function and (roughly) its own derivative, substitute the inner function as u.
Trigger: You see f(g(x)) · g'(x). Set u = g(x), du = g'(x) dx. The integral collapses.
Example: ∫ x√(x² + 1) dx
Look — x dx is half the derivative of x² + 1. Set u = x² + 1, du = 2x dx, so x dx = du/2.
Step count without substitution: You'd have to expand, attempt partial fractions, fail — several wasted minutes. With substitution: 4 lines.
Many trig integrals are unsolvable directly but become trivial after an identity. The two most common reductions:
Half-angle forms:
1 - cos x = 2 sin²(x/2) — this is the key identity for the NDA 2025 question on ∫ 1/(1-cos x) dx1 + cos x = 2 cos²(x/2)sin²x = (1 - cos 2x)/2 — converts ∫sin²x dx into a simple standard formProtocol: Whenever you see 1 - cos x, 1 + cos x, sin²x, cos²x in the denominator or as the main integrand — apply the half-angle or double-angle identity immediately before doing anything else.
Used when the integrand is a product of two "unrelated" functions. The formula:
ILATE priority for choosing u: Inverse trig → Logarithm → Algebraic → Trigonometric → Exponential. Whatever comes first in ILATE is your u.
NDA rarely goes deeper than one application of integration by parts — they don't test iterated IBP or reduction formulae at this level. So if you're doing IBP twice on an NDA question, reconsider whether substitution works faster.
The form ∫ 1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + C catches many candidates off guard because of the 1/a coefficient. The trap option always places tan⁻¹(x/a) + C (missing the 1/a) or 2 tan⁻¹(x/a) + C. Always verify: differentiate your answer and check it matches the integrand.
For a = 2: ∫ 1/(x² + 4) dx = (1/2) tan⁻¹(x/2) + C. Not tan⁻¹(x/2), not (1/4) tan⁻¹(x/2). The 1/a factor is 1/2.
The four trig integrals (sin, cos, sec², cosec²) alternate signs in a specific pattern. Write:
Pattern: negative, positive, positive, negative. Or remember "sin starts negative, cos starts positive, sec² positive, cosec² negative." Cross-check in 2 seconds by noticing: differentiating sin gives +cos (not -cos), so integrating goes opposite → -cos. This eliminates sign errors on option lists. Standard time to recall: 15s without this pattern; under 3s with it.
Before writing a single line, scan the integrand for this: is there a composite function AND something proportional to its inner derivative also present? If yes, substitution will work. Specifically:
x and x² + k → let u = x² + ksin x and cos x in numerator/denominator → let u = whichever is in the denominatorf'(x)/f(x) → that's d/dx [ln|f(x)|], answer is ln|f(x)| + CThis scan takes under 10 seconds and saves you from attempting integration by parts on a substitution problem — which would cost 3+ minutes.
Any time you see 1 - cos x or 1 + cos x (especially in the denominator), fire the half-angle identity immediately without thinking:
1 - cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2)
The resulting integrand will always reduce to a cosec²(x/2) or sec²(x/2) form, giving a cot(x/2) or tan(x/2) answer. The only remaining step is the chain rule coefficient: ∫cosec²(x/2) dx. Since the inner function is x/2, its derivative is 1/2, so you divide by 1/2 (i.e., multiply by 2):
∫cosec²(x/2) dx = -2cot(x/2) + C
The 2025 NDA question is solved in 3 steps using this reflex. Without it: 8+ steps.
After getting an answer, differentiate it mentally and check it matches the original integrand. This takes 15-30 seconds and eliminates all sign errors and coefficient errors. In a 4-option MCQ, the wrong options are typically the answer with a wrong sign or a wrong coefficient — one fast verification step separates full marks from zero on every integration question. Effective step-count saved: you avoid picking the "mirror" wrong answer, which appears in roughly 3 of 4 option lists on NDA integration questions.
For integration by parts ∫u dv = uv - ∫v du, you need to pick u instantly. Use ILATE in order:
Inverse trig → Log → Algebra → Trig → Exponential
The first function present in this list becomes u. Example: ∫x eˣ dx — x is Algebraic, eˣ is Exponential. A comes before E, so u = x, dv = eˣ dx. Result: x eˣ - eˣ + C. This decision takes under 5 seconds with ILATE. Without a rule, candidates spend 30-60 seconds trying both choices to see which simplifies.
When an NDA integration question appears, run this decision tree:
Step 1 — Is it a direct standard form? Look at the integrand. Does it match xⁿ, sin x, cos x, 1/(x²+a²) etc. immediately? If yes, write the answer in under 20 seconds.
Step 2 — Is there a trig identity hiding? If you see sin²x, cos²x, 1 - cos x, 1 + cos x — apply the relevant identity first. The integral will collapse to a standard form.
Step 3 — Substitution scan. Is there a composite function with its derivative (or proportional derivative) present? Set u = inner function, convert, integrate.
Step 4 — Integration by parts. Is it a product of two unrelated function types? Apply ILATE, run one cycle of IBP.
If none of the above work within 30 seconds: look at the answer options. Work backwards — differentiate each option and see which matches the integrand. This reverse-check takes about 60 seconds and guarantees the correct answer on any standard NDA question without requiring you to execute the forward method at all.
Never guess without at least a sign-check on the options.
Why this question: This is a 2025 NDA question that tests the half-angle identity reflex — if you don't see 1 - cos x = 2sin²(x/2) immediately, the problem looks impossible.
Solving path: Recognise 1 - cos x = 2sin²(x/2). Then y = 1/(2sin²(x/2)) = (1/2)cosec²(x/2). Integrate: (1/2) × (-2cot(x/2)) + C = -cot(x/2) + C. The factor of 2 in the antiderivative of cosec²(x/2) comes from the chain rule (inner derivative is 1/2, so you divide by 1/2). Answer: option B.
Why this question: Tests the 1/(x² + a²) standard form — the most commonly confused formula because of the 1/a coefficient that many candidates forget.
Solving path: Match to ∫ 1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + C with a = 2. Result: (1/2)tan⁻¹(x/2) + C. Verify: differentiate (1/2)tan⁻¹(x/2) using chain rule: (1/2) × 1/(1 + x²/4) × (1/2) = (1/2) × 4/(4 + x²) × (1/2) = 1/(x² + 4). Confirmed. Answer: option A.
Why this question: Classic substitution problem. Recognising x dx as half of d(x²+1) is the key insight — the entire problem resolves in 4 lines.
Solving path: Let u = x² + 1, du = 2x dx, so x dx = du/2. Integral becomes ∫√u × (du/2) = (1/2)∫u^(1/2) du = (1/2) × (2/3)u^(3/2) + C = (1/3)(x²+1)^(3/2) + C. Answer: option A.
Why this question: Tests the double-angle identity for sin²x — a very common NDA integrand that cannot be handled directly.
Solving path: Apply sin²x = (1 - cos 2x)/2. Then ∫sin²x dx = ∫(1 - cos 2x)/2 dx = (1/2)∫1 dx - (1/2)∫cos 2x dx = x/2 - (1/2)(sin 2x / 2) + C = x/2 - sin(2x)/4 + C. Answer: option A. The trap is option B (wrong sign on the sin 2x term).
Why this question: Straightforward definite integral but included to show the evaluate-and-substitute process — also confirms that + C cancels in definite integrals.
Solving path: ∫(2x + 3) dx = x² + 3x + C. Evaluate from 1 to 4: [4² + 3(4)] - [1² + 3(1)] = [16 + 12] - [1 + 3] = 28 - 4 = 24. Answer: option B. Note C cancels — you never need C for definite integrals.
Dropping the + C entirely. In indefinite integration, omitting + C is technically an incomplete answer. In a MCQ context where options differ by the constant, you might also find yourself confused. Always write + C — it reinforces the habit.
Sign error on ∫cosec²x dx. The answer is -cot x + C, not +cot x + C. The positive form appears as a trap option in almost every NDA question involving cosec². Differentiate -cot x to verify: d/dx(-cot x) = cosec²x. Correct.
Forgetting the 1/a factor in ∫ 1/(x²+a²) dx. The formula is (1/a)tan⁻¹(x/a) + C. The bare tan⁻¹(x/a) + C (without 1/a) is always one of the wrong options. The 30-second differentiation check catches this every time.
Applying IBP to a substitution problem. If you see ∫ x f(x²) dx, that x is screaming "I am the derivative of x²". Substitution is always faster here. IBP on the same problem gives a longer path and more chances to slip.
Forgetting the chain-rule factor when integrating composite functions. ∫ cosec²(x/2) dx is NOT -cot(x/2) + C. The inner function is x/2 with derivative 1/2, so you multiply the antiderivative by 1/(1/2) = 2: result is -2cot(x/2) + C. Missing this factor of 2 costs you the exact answer.
Using ∫ sin²x dx = -cos²x / (2 sin x) + C (wrong manipulation). This incorrect algebraic manipulation appears when students try to treat sin²x as if differentiation rules apply directly. The correct approach is always the identity sin²x = (1 - cos 2x)/2 first. There is no shortcut that bypasses the identity.