Integration is just differentiation run in reverse — instead of finding the slope, you rebuild the original function. Add a few clever tricks for areas under curves and you have one of the most predictable and high-scoring parts of NDA Maths. Together, indefinite and definite integration carry roughly 8–12 marks every year, and the patterns repeat faithfully.
Why Integration Is a Guaranteed Scoring Zone
Calculus makes up nearly a fifth of the NDA Maths paper, and integration is the half of it that students find most reliable once they learn the patterns. Unlike geometry, where a single tricky figure can eat ten minutes, an integration problem is usually a matter of spotting the right method and turning the handle. The work is mechanical once the method is chosen, which means fewer surprises and far less guesswork under exam pressure.
It also helps that integration links directly to topics you already know. Areas, volumes and averages all reduce to a definite integral, so the effort you put in here quietly pays off across several other questions in the paper. Think of it as a skill that compounds rather than a chapter you cram and forget.
The examiners reuse a small toolkit year after year: standard formulas, a substitution, integration by parts, or a definite-integral property. There is rarely anything exotic. That predictability is exactly why we tell Cavalier students to treat this chapter as a marks bank — invest a few focused hours and you can almost guarantee yourself the questions.
Integration is the reverse of differentiation, so your differentiation table is half your integration table. If you know that the derivative of sin x is cos x, you instantly know that the integral of cos x is sin x + C.
What Integration Really Means
An indefinite integral answers a simple question: which function, when differentiated, gives me back the one I started with? That original function is called the antiderivative or primitive.
If d/dx [F(x)] = f(x), then ∫ f(x) dx = F(x) + C.
Here f(x) is the integrand, dx tells us the variable, and C is the constant of integration.
Why the +C? Because differentiating any constant gives zero, infinitely many functions share the same derivative — they differ only by a constant. So an indefinite integral is really a whole family of curves stacked vertically, not a single answer.
Forgetting the + C in an indefinite integral is the single most common slip. Examiners often include an option with and without the constant to catch hurried students. Always write it.
Standard Integrals You Must Memorise
These are the building blocks. Learn them cold and most NDA integration questions become one-line problems.
∫ xn dx = xn+1/(n+1) + C, for n ≠ −1
∫ 1/x dx = log|x| + C
∫ ex dx = ex + C
∫ ax dx = ax/log a + C
∫ sin x dx = −cos x + C
∫ cos x dx = sin x + C
∫ sec2x dx = tan x + C
∫ cosec2x dx = −cot x + C
Two of these trip people up. The power rule fails when n = −1, because you would divide by zero; that gap is exactly why ∫ 1/x dx = log|x| + C exists as a special case. And the sine integral picks up a minus sign — a favourite trap.
Sign rule for trig: integrating sin gives −cos, and integrating cosec2 gives −cot. The two functions that begin with the same letters as the answer take the minus sign.
The Basic Rules of Integration
Integration is linear, which makes long expressions easy to break apart.
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
∫ k·f(x) dx = k ∫ f(x) dx, where k is a constant
So a polynomial like 3x2 + 5x − 7 is integrated term by term. There is, however, no product rule and no quotient rule for integration the way there is for differentiation — products and quotients usually need substitution or by-parts instead.
You cannot integrate a product by integrating each factor separately. ∫ (sin x · cos x) dx is not (−cos x)(sin x). That mistake loses easy marks every year.
Integration by Substitution
When the integrand contains a function and its derivative, substitution turns a messy integral into a standard one. The idea is to replace an inner expression with a new variable t.
Put t = g(x), so that dt = g′(x) dx. Then ∫ f(g(x))·g′(x) dx = ∫ f(t) dt.
A hugely useful special case is worth memorising on its own, because it appears constantly:
∫ f′(x)/f(x) dx = log|f(x)| + C
Whenever the numerator is the derivative of the denominator, the answer is simply log of the denominator.
For example, ∫ tan x dx = ∫ sin x/cos x dx. The top is (almost) the derivative of the bottom, so it becomes −log|cos x| + C, which is also written as log|sec x| + C.
Scan for a chunk whose derivative also sits in the integrand. If you spot it, substitution will almost certainly work. Practise recognising this pairing — it is the most-tested integration skill in NDA.
Integration by Parts
When you must integrate a product of two different kinds of function — say x times ex, or x times log x — substitution will not help. Integration by parts is the tool.
∫ u·v dx = u ∫ v dx − ∫ [ (du/dx) · ∫ v dx ] dx
Choose u (the part to differentiate) using ILATE: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential — whichever comes first becomes u.
The ILATE order tells you which factor to call u so the remaining integral gets simpler, not harder. For ∫ x·ex dx, Algebraic (x) beats Exponential (ex), so u = x. Differentiating x gives 1, which kills the polynomial and leaves an easy integral.
For a lone log or inverse-trig function like ∫ log x dx, write it as 1 × log x and apply by-parts with v = 1. The answer is x·log x − x + C.
Definite Integrals and the Fundamental Theorem
A definite integral has limits and gives a number, not a family of functions — there is no +C. It measures the signed area under the curve y = f(x) between x = a and x = b.
Fundamental Theorem of Calculus:
∫ab f(x) dx = F(b) − F(a), where F is any antiderivative of f.
So you integrate as usual, then plug in the upper limit minus the lower limit. The constant C cancels out, which is why definite integrals never carry one.
If you use substitution in a definite integral, you must change the limits too — convert the x-limits into t-limits. Forgetting this and plugging x-values into a t-expression is a classic error.
Powerful Properties of Definite Integrals
These properties let you solve integrals that look impossible to compute directly. They are heavily tested in NDA because the trick collapses a hard problem into a trivial one.
1. ∫ab f(x) dx = ∫ab f(a+b−x) dx
2. ∫0a f(x) dx = ∫0a f(a−x) dx
3. ∫−aa f(x) dx = 2∫0a f(x) dx if f is even; = 0 if f is odd
The even/odd property is the biggest time-saver. If f(−x) = −f(x) over a symmetric interval, the integral is simply zero — the area on the left cancels the area on the right. So ∫−ππ sin x dx = 0 with no calculation at all.
If you see symmetric limits like −a to a, immediately test whether the integrand is odd or even. Half the time the answer is zero and you save two minutes.
Worked Example: Substitution in Action
Let us walk through a typical NDA-level integral that rewards method over memory.
Evaluate ∫ 2x · ex2 dx.
Notice how the 2x in front was not random — it was exactly the derivative of the exponent. Recognising that pairing is the whole skill. Whenever a stray factor matches the derivative of an inner function, substitution dissolves the problem.
Always substitute back to the original variable at the end of an indefinite integral. An answer left in terms of t is incomplete.
Common Mistakes That Cost Easy Marks
Most integration errors are careless rather than conceptual. Knowing the usual traps is half the battle.
- Dropping the +C in indefinite integrals — an automatic option-trap.
- Wrong trig sign — writing ∫ sin x dx as +cos x instead of −cos x.
- Not changing limits after substitution in a definite integral.
- Inventing a product rule — integration has none; use by-parts.
- Ignoring the n = −1 case — ∫ 1/x dx is log|x|, never x0/0.
Students often write ∫ 1/x2 dx incorrectly. Here n = −2, so the power rule does apply: the answer is x−1/(−1) = −1/x + C. Only n = −1 is the exception.
Previous-Year Style Question
Here is a question in the exact flavour the NDA exam loves — a definite integral cracked open by a clever property rather than brute force.
Q. Evaluate ∫0π/2 sin x / (sin x + cos x) dx.
Answer: Call the integral I. Using the property ∫0a f(x) dx = ∫0a f(a−x) dx with a = π/2 swaps sin and cos, giving a second integral with cos x / (cos x + sin x). Adding the original I and this new form, the numerators combine to (sin x + cos x), which cancels the denominator, leaving 2I = ∫0π/2 1 dx = π/2. Therefore I = π/4.
Notice that we never actually integrated the original messy fraction. The symmetry property did all the heavy lifting — this is precisely the kind of shortcut the NDA paper rewards.
Quick Revision Before the Exam
Run through this checklist the night before and integration questions will feel routine.
- Indefinite integral = antiderivative + C; definite integral = a number (no C).
- Power rule: ∫ xn dx = xn+1/(n+1) + C, except n = −1 which gives log|x|.
- ∫ f′(x)/f(x) dx = log|f(x)| + C — spot derivative-over-function.
- Substitution for function-and-its-derivative; by-parts (ILATE) for products.
- Fundamental Theorem: ∫ab f dx = F(b) − F(a).
- Odd function over −a to a integrates to zero; even function doubles 0 to a.
- Change the limits when substituting in a definite integral.
Practise five mixed integrals daily for a fortnight and this chapter will become one of your most dependable scorers in the entire paper. At Cavalier, students who drill these patterns rarely lose a single integration mark in the actual NDA exam.
Frequently asked questions
What is the difference between indefinite and definite integration?
An indefinite integral has no limits and gives a family of functions with a +C constant, representing all antiderivatives. A definite integral has lower and upper limits and gives a single number, usually the signed area under the curve between those limits.
How many marks does integration carry in the NDA exam?
Indefinite and definite integration together typically account for about 8 to 12 marks in each NDA Maths paper. Because the question patterns repeat, it is one of the most reliable scoring areas in the syllabus.
When should I use substitution versus integration by parts?
Use substitution when the integrand contains a function together with its derivative, so you can replace the inner part with t. Use integration by parts when you must integrate a product of two unrelated functions, choosing u by the ILATE order.
Why do we add +C in indefinite integrals?
Because differentiating any constant gives zero, infinitely many functions share the same derivative, differing only by a constant. The +C represents this entire family of antiderivatives. Definite integrals do not need it because the constant cancels out.
What is the easiest way to evaluate a definite integral with symmetric limits?
Check whether the integrand is odd or even. If it is odd, the integral from minus-a to a is zero; if it is even, it equals twice the integral from zero to a. This often turns a hard problem into a one-line answer.
Related NDA Maths topics
Want a teacher to walk you through NDA Maths?
Cavalier's NDA batches break every topic into classroom sessions with daily practice, tests and doubt-clearing.