Why technique selection is the whole skill
Consider ∫x e^x dx and ∫x e^(x²) dx. They look almost identical. The first needs integration by parts; the second is a substitution, because 2x is proportional to the derivative of x². Choosing the wrong one on either turns a one-line answer into an unbounded mess.
That is why practising integration by topic - a page of substitutions, then a page of by-parts - trains the wrong thing. The exam does not label the technique, and a mixed set is the only practice that develops the recognition the paper actually tests.
How JEE actually asks Integration
NTA does not publish chapter-wise weightage, so any figure online is a coaching estimate from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
Past papers show a consistent split. Indefinite integrals requiring substitution or by parts supply the bulk. Definite integrals frequently exploit the symmetry properties rather than requiring an antiderivative at all - and those are the questions that look hardest and are actually fastest. Area under and between curves rounds it out, usually as a short application.
Key concepts, compressed
- ✓Integration is the reverse of differentiation, so every derivative result gives a standard integral for free.
- ✓The constant of integration is part of the answer for an indefinite integral, and in differential-equation questions it carries the initial condition.
- ✓Substitution reverses the chain rule: it works when the integrand contains f(g(x)) together with g′(x).
- ✓Integration by parts reverses the product rule: ∫u dv = uv − ∫v du, with u chosen by ILATE.
- ✓A definite integral is a number, evaluated by the fundamental theorem as F(b) − F(a).
- ✓The symmetry properties of definite integrals often make an antiderivative unnecessary.
Results you need before attempting the questions
| Integral | Result | Note |
|---|---|---|
| ∫xⁿ dx | x^(n+1)/(n+1) + C | n ≠ −1 |
| ∫(1/x) dx | ln|x| + C | the n = −1 case |
| ∫eˣ dx | eˣ + C | |
| ∫sin x dx | −cos x + C | minus sign |
| ∫cos x dx | sin x + C | |
| ∫sec²x dx | tan x + C | |
| ∫1/(1 + x²) dx | tan⁻¹x + C | |
| ∫1/√(1 − x²) dx | sin⁻¹x + C | |
| ∫tan x dx | ln|sec x| + C | = −ln|cos x| + C |
| By parts | ∫u dv = uv − ∫v du | u by ILATE |
| Symmetry property | ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx | |
| Odd / even over [−a, a] | 0 / 2∫₀ᵃ f dx | |
| Area between curves | ∫ (upper − lower) dx |
The five mistakes that cost the most marks
- ✓Omitting the constant of integration. It is a standard deduction, and in differential-equation questions it is the entire answer.
- ✓Failing to change the limits after substituting in a definite integral. Either convert the limits or substitute back before evaluating - never mix the two.
- ✓Choosing u badly in integration by parts. ILATE exists precisely to stop you differentiating the exponential and integrating the logarithm.
- ✓Computing area as a plain definite integral when the curve crosses the axis. The integral counts area below the axis as negative; area does not.
- ✓Applying ∫xⁿ dx = x^(n+1)/(n+1) at n = −1. That case gives ln|x|, and the absolute value matters for negative x.
Practice set 1: standard integrals
1. Evaluate ∫x³ dx.
x⁴/4 + C. The power rule for integration raises the exponent by one and divides by the new exponent - exactly reversing differentiation. Check any integration by differentiating your answer: d/dx(x⁴/4) = x³, which recovers the integrand.
2. Evaluate ∫(1/x) dx.
ln|x| + C. This is the exception to the power rule, since applying it with n = −1 would divide by zero. The absolute value is required because 1/x is defined for negative x while ln x is not - and omitting it is a genuine error, not a formality.
3. Evaluate ∫eˣ dx.
eˣ + C. The exponential function is its own derivative and therefore its own integral, which is what makes it central to differential equations. Note that ∫e^(kx) dx = e^(kx)/k + C - the constant divides on integrating, whereas it multiplies on differentiating.
4. Evaluate ∫sin x dx.
−cos x + C, with the minus sign. This is the reverse of d/dx(cos x) = −sin x. The companion result ∫cos x dx = sin x + C has no minus sign, and swapping the two signs is one of the most frequent slips in the chapter.
5. Evaluate ∫sec²x dx.
tan x + C, directly reversing d/dx(tan x) = sec²x. It is worth reading the standard derivative table backwards once and treating it as an integral table - roughly half of the standard integrals come free that way.
6. Evaluate ∫1/(1 + x²) dx.
tan⁻¹x + C. Recognising this form is high-value, because many integrals reduce to it after completing the square in the denominator. The general version is ∫1/(a² + x²) dx = (1/a)tan⁻¹(x/a) + C.
7. Evaluate ∫1/√(1 − x²) dx.
sin⁻¹x + C. Note how similar this looks to question 6 while giving a completely different function - the square root and the minus sign are what distinguish them. Misreading one for the other is a standard trap, so check the sign and the root before writing anything.
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Practice set 2: substitution
8. Evaluate ∫2x/(1 + x²) dx.
ln(1 + x²) + C. Substituting u = 1 + x² gives du = 2x dx, and the integral becomes ∫du/u = ln|u| + C. The signal was that the numerator is exactly the derivative of the denominator - whenever you see that, the answer is the logarithm of the denominator. No absolute value is needed here since 1 + x² is always positive.
9. Evaluate ∫sin x cos x dx.
sin²x/2 + C. With u = sin x, du = cos x dx, giving ∫u du = u²/2. You could equally substitute u = cos x to get −cos²x/2 + C, or use the double-angle identity to get −cos2x/4 + C. All three differ only by a constant, which is absorbed into C - a genuinely useful thing to have seen before an exam.
10. Evaluate ∫x e^(x²) dx.
e^(x²)/2 + C. Substituting u = x² gives du = 2x dx, so x dx = du/2 and the integral becomes (1/2)∫e^u du. Compare with ∫x eˣ dx in the next set, which looks similar but has no derivative relationship and needs integration by parts instead. Spotting the difference is the point.
11. Evaluate ∫(2x + 3)⁵ dx.
(2x + 3)⁶/12 + C. Substituting u = 2x + 3 gives du = 2dx, so the integral is (1/2)∫u⁵ du = u⁶/12. The general result for a linear inner function is ∫(ax + b)ⁿ dx = (ax + b)^(n+1)/(a(n+1)) + C - dividing by the coefficient a is the step most often forgotten.
12. What signals that substitution is the right technique?
The integrand contains a composite function together with something proportional to the derivative of the inner function. The word proportional matters: ∫x e^(x²) dx works even though the derivative of x² is 2x rather than x, because the constant factor can be pulled out. Without that relationship, substitution leads nowhere.
13. Evaluate ∫tan x dx.
ln|sec x| + C, equivalently −ln|cos x| + C. Write tan x as sin x/cos x and substitute u = cos x, giving du = −sin x dx and the integral −∫du/u = −ln|u|. The two forms are identical since −ln|cos x| = ln|1/cos x| = ln|sec x|, and either is acceptable.
Practice set 3: integration by parts
14. What does ILATE stand for and what is it used for?
Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential - the priority order for choosing which factor to call u in ∫u dv = uv − ∫v du. Whichever type comes first in that list becomes u and gets differentiated. The logic is that inverse and logarithmic functions simplify dramatically on differentiation, while exponentials never do.
15. Evaluate ∫x eˣ dx.
eˣ(x − 1) + C. By ILATE, the algebraic x comes before the exponential, so u = x and dv = eˣ dx. Then ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C. Had you chosen u = eˣ instead, the remaining integral would contain x²/2 - worse than where you started, which is exactly what ILATE prevents.
16. Evaluate ∫x sin x dx.
−x cos x + sin x + C. Take u = x and dv = sin x dx, so v = −cos x. Then the result is −x cos x + ∫cos x dx = −x cos x + sin x + C. Verify by differentiating: −cos x + x sin x + cos x = x sin x, which recovers the integrand. That check takes ten seconds and catches every sign error.
17. Evaluate ∫ln x dx.
x ln x − x + C. There appears to be only one function, but write it as ∫(ln x)(1) dx with u = ln x and dv = dx. Then v = x and the result is x ln x − ∫x(1/x) dx = x ln x − ∫dx = x ln x − x + C. Treating a lone logarithm or inverse trigonometric function as a product with 1 is a standard move worth remembering.
18. How would you evaluate ∫x² eˣ dx?
By applying integration by parts twice. The first application reduces x² to 2x, leaving ∫2x eˣ dx, and the second reduces that to a constant. The result is eˣ(x² − 2x + 2) + C. In general, an integrand with xⁿ times an exponential needs n applications, each one lowering the power by one.
19. What happens if you choose u badly in integration by parts?
The remaining integral becomes harder rather than easier. Choosing u = eˣ in ∫x eˣ dx produces ∫(x²/2)eˣ dx - a higher power than you began with. If a by-parts application leaves you worse off, do not push on: reverse the choice of u and dv.
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Practice set 4: definite integrals
20. Evaluate ∫₀¹ x² dx.
1/3. The antiderivative is x³/3, so by the fundamental theorem the value is 1³/3 − 0³/3 = 1/3. Note that a definite integral is a number, so no constant of integration appears - it cancels in the subtraction, which is why it can be dropped from the start here.
21. Evaluate ∫₀^π sin x dx.
2. The antiderivative is −cos x, giving −cos π − (−cos 0) = −(−1) + 1 = 2. Watch the double negative, which is where this specific integral is most often mishandled. Geometrically it is the area of one arch of the sine curve, which is exactly 2.
22. State the most useful symmetry property of definite integrals.
∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx. Adding the integral to its transformed version frequently makes the difficult part cancel, converting an intractable integrand into a constant. JEE constructs questions around this deliberately - an integral that looks impossible is often signalling that this property is intended.
23. What is ∫₋ₐᵃ f(x)dx when f is odd?
Zero. An odd function satisfies f(−x) = −f(x), so the contributions from the two halves of a symmetric interval cancel exactly. Checking whether an integrand is odd should be your first move on any symmetric-interval question - it can end the problem in one line, and x³, sin x and tan x are all odd.
24. What is ∫₋ₐᵃ f(x)dx when f is even?
2∫₀ᵃ f(x)dx. An even function satisfies f(−x) = f(x), so both halves contribute equally. This halves the work rather than eliminating it. Note that a general function is neither odd nor even, and splitting it into odd and even parts is sometimes the intended route.
25. What must you do to the limits when substituting in a definite integral?
Convert them to the new variable, or else substitute back to the original variable before evaluating. Doing neither - computing an antiderivative in u and then applying the original x limits - produces a wrong number with no warning sign. Converting the limits is usually cleaner because it avoids the back-substitution entirely.
Practice set 5: area under and between curves
26. Find the area under y = x² from x = 0 to x = 3.
9 square units. The area is ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9. Because the curve lies entirely above the axis on this interval, the definite integral and the area coincide - which is not true in general, as question 28 shows.
27. How do you find the area between two curves?
Integrate the difference, upper curve minus lower curve, between their points of intersection: ∫(upper − lower) dx. Find the intersection points first by equating the two functions, since those are the limits. If the curves cross within the region, the roles of upper and lower swap and the region must be split there.
28. Why is the area under y = sin x from 0 to 2π not equal to the definite integral?
Because the curve is below the axis on the second half. The definite integral is 2 + (−2) = 0, since the negative contribution cancels the positive. The area is 2 + |−2| = 4. Whenever a curve crosses the axis inside the interval, split at the crossing points and add the magnitudes.
29. Evaluate ∫₀^(π/2) sin x dx.
1. The antiderivative is −cos x, so the value is −cos(π/2) + cos 0 = 0 + 1 = 1. This is exactly half the value from question 21, which the symmetry of the sine arch about x = π/2 makes obvious - a useful sanity check on both answers.
30. Find the area enclosed between y = x and y = x² from x = 0 to x = 1.
1/6 square units. The two curves meet at x = 0 and x = 1, and on that interval x ≥ x², so x is the upper curve. The area is ∫₀¹(x − x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6. Getting the order of subtraction wrong gives −1/6, and a negative area is always a signal that the curves have been swapped.
How to study this chapter efficiently
- ✓Practise mixed sets from the beginning. Practising by technique trains recognition of the section heading, not of the integrand.
- ✓Before integrating, ask one question: does the integrand contain a function and its derivative? If yes, substitute. If it is a product of unrelated types, use parts.
- ✓Differentiate your answer to check it. This is the single most effective error-catching habit in the chapter and takes seconds.
- ✓For any definite integral over a symmetric interval, test for odd or even symmetry first.
- ✓Memorise the symmetry property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx and actively look for chances to use it. Several JEE questions per year are built on it.
- ✓For area problems, always sketch. It tells you which curve is upper and whether the region crosses the axis - both invisible from the algebra.
Turn this into active practice
The recognition this chapter needs cannot be built from worked examples, because a solved integral has already had its technique chosen. You learn the choice by facing an unlabelled integrand, picking wrong, and hitting a dead end - which is unpleasant on paper and cheap in practice.
The JEE Integration quiz on QUFF generates fresh questions across standard forms, substitution, by parts, definite-integral properties and areas, marks them instantly and explains each answer. Do a deliberately mixed set, and for each one you get wrong record whether the technique choice or the execution failed.
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