JEE Maths — Definite Integrals Complete Chapter Guide

Chapter Overview & Weightage

Definite Integrals is one of the most reliable scoring chapters in JEE Maths. Every year, without exception, you’ll see 1-2 questions from this chapter — and the good news is that most of them follow predictable patterns.

Weightage: 6-8% of JEE Maths paper. That’s roughly 2-3 questions in JEE Main (8-12 marks). In JEE Advanced, this chapter pairs with Area Under Curves and sometimes Differential Equations for multi-concept problems.

Year	JEE Main Questions	Marks	Key Topics Asked
2024	2	8	Property-based evaluation, Walli’s formula
2023	2	8	Leibniz rule, symmetric property
2022	3	12	Reduction formula, $\int_0^{\pi/2}$ type
2021	2	8	King’s property, definite integral as limit
2020	2	8	Even/odd function property, substitution

The pattern is clear: property-based questions dominate. If you master the 8 standard properties, you can solve 70% of JEE Main definite integral questions without actually integrating.

Key Concepts You Must Know

Prioritized by how often they appear in PYQs:

Tier 1 — Appears Almost Every Year

King’s property: $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$
Even/odd function property over symmetric limits $[-a, a]$
$\int_0^{\pi/2} f(\sin x)\,dx = \int_0^{\pi/2} f(\cos x)\,dx$
Definite integral as limit of a sum (Riemann sum form)

Tier 2 — High Frequency

Walli’s formula for $\int_0^{\pi/2} \sin^m x \cos^n x\,dx$
Leibniz rule for differentiating under the integral sign
Periodic function property: $\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx$
Breaking limits when integrand has modulus or discontinuity

Tier 3 — Advanced / JEE Advanced Focus

Reduction formulas ( $I_n$ type recurrences)
Integration of functions with floor/fractional part
Comparison of definite integrals without evaluating

Important Formulas

P1 — Order reversal:

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx

P2 — Splitting limits (most used for modulus questions):

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

P3 — King’s property (exam favourite):

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

P4 — For limits $[0, a]$ :

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

P5 — Even/odd over $[-a, a]$ :

\int_{-a}^a f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}

P6 — Trigonometric (extremely high yield):

\int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx

P7 — Periodic functions ( $T$ = period):

\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx

P8 — Half-period for symmetric functions:

\int_0^{2a} f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx & \text{if } f(2a-x) = f(x) \\ 0 & \text{if } f(2a-x) = -f(x) \end{cases}

For $I_{m,n} = \int_0^{\pi/2} \sin^m x \cos^n x\,dx$ :

I_{m,n} = \frac{(m-1)!! \cdot (n-1)!!}{(m+n)!!} \cdot K

where $K = \pi/2$ if both $m$ and $n$ are even, else $K = 1$ .

When to use: Any integral of the form $\int_0^{\pi/2} \sin^m x \cos^n x\,dx$ where direct computation is messy. Saves 3-4 minutes per question.

If $F(x) = \int_{g(x)}^{h(x)} f(t)\,dt$ , then:

F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)

When to use: Whenever you see “find $\frac{d}{dx}$ of a definite integral” or limits that are functions of $x$ . Also appears in limit problems: $\lim_{x \to a} \frac{\int_a^x f(t)\,dt}{x-a} = f(a)$ .

\int_a^b f(x)\,dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{r=0}^{n-1} f\!\left(a + r\cdot\frac{b-a}{n}\right)

For $[0,1]$ , the standard form is:

\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx

When to use: Whenever you see a limit of a sum involving $n$ and $r/n$ terms. Convert to integral, then evaluate.

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (January, Shift 2)

Question: Evaluate $\displaystyle\int_0^\pi \frac{x \sin x}{1 + \cos^2 x}\,dx$

Solution:

Let $I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x}\,dx$ .

Apply King’s property (P3): replace $x$ with $\pi - x$ .

I = \int_0^\pi \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}\,dx = \int_0^\pi \frac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx

Add both expressions:

2I = \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x}\,dx

2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x}\,dx

Substitute $t = \cos x$ , $dt = -\sin x\,dx$ . Limits: $x=0 \Rightarrow t=1$ , $x=\pi \Rightarrow t=-1$ .

2I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^{1} \frac{dt}{1+t^2} = \pi \big[\arctan t\big]_{-1}^{1}

2I = \pi\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \frac{\pi^2}{2}

\boxed{I = \frac{\pi^2}{4}}

Why King’s property works here: The integrand has $x$ multiplied by a function of $\sin x$ and $\cos^2 x$ . Whenever you see $x \cdot g(\sin x, \cos x)$ with limits $[0, \pi]$ or $[0, a]$ , King’s property is almost always the right move — it turns the uncomfortable $x$ into the constant $\pi$ .

PYQ 2 — JEE Main 2023 (April, Shift 1)

Question: $\displaystyle\lim_{n \to \infty} \frac{1}{n}\left(\frac{1}{\sqrt{n}} + \frac{1}{\sqrt{2n}} + \frac{1}{\sqrt{3n}} + \cdots + \frac{1}{\sqrt{n^2}}\right)$

Solution:

Rewrite the sum:

= \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{rn}} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{n} \cdot \sqrt{r}}

= \lim_{n \to \infty} \frac{1}{n^{3/2}} \sum_{r=1}^{n} \frac{1}{\sqrt{r}}

Hmm, that doesn’t match standard form. Let’s rewrite carefully:

= \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{r/n} \cdot n} \cdot n

Actually: $\frac{1}{\sqrt{rn}} = \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{r}}$ , so:

= \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{n} \cdot \sqrt{r}} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{r/n}} \cdot \frac{1}{n}

Wait — standard form: $\frac{1}{n}\sum f(r/n) \to \int_0^1 f(x)\,dx$ . Here:

\frac{1}{\sqrt{rn}} = \frac{1}{n} \cdot \frac{1}{\sqrt{r/n}}

So the sum becomes $\frac{1}{n}\sum_{r=1}^n \frac{1}{\sqrt{r/n}}$ , which converts to:

\int_0^1 \frac{1}{\sqrt{x}}\,dx = \int_0^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_0^1 = 2

\boxed{\text{Answer} = 2}

PYQ 3 — JEE Main 2022 (June, Shift 1)

Question: $\displaystyle\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 3^x}\,dx$

Solution:

Limits are symmetric: $[-\pi/2, \pi/2]$ . Check if we can use the split trick.

Let $I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 3^x}\,dx$ .

Apply $x \to -x$ :

I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{1 + 3^{-x}}\,dx = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{\frac{1+3^x}{3^x}}\,dx = \int_{-\pi/2}^{\pi/2} \frac{3^x \cos^2 x}{1 + 3^x}\,dx

Add:

2I = \int_{-\pi/2}^{\pi/2} \cos^2 x \cdot \frac{1 + 3^x}{1 + 3^x}\,dx = \int_{-\pi/2}^{\pi/2} \cos^2 x\,dx

Since $\cos^2 x$ is even:

2I = 2\int_0^{\pi/2} \cos^2 x\,dx = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}

\boxed{I = \frac{\pi}{4}}

The trick here: Any integral of the form $\int_{-a}^{a} \frac{f(x)}{1 + a^x}\,dx$ where $f(x)$ is even equals $\int_0^a f(x)\,dx$ . This is a JEE favourite. The $3^x$ in denominator looks intimidating but vanishes completely with this method.

Difficulty Distribution

For JEE Main, here’s roughly what you’ll face:

Difficulty	% of Questions	What It Looks Like
Easy	35%	Direct property application (King’s, even/odd), standard Walli’s
Medium	50%	Two-step property chains, Leibniz rule, limit-of-sum conversion
Hard	15%	Reduction formula recursion, mixed modulus + periodic, JEE Advanced multi-step

In JEE Main, almost no question requires heavy computation if you recognise the correct property. The difficulty is in pattern recognition, not calculation. Spend your revision time drilling property identification, not computing antiderivatives.

Expert Strategy

Week 1 — Build the property toolkit. Write all 8 properties on one page. For each, write 2 PYQs that use it. Don’t move forward until you can look at a definite integral and immediately know which property applies.

Week 2 — Walli’s and Leibniz. These two together cover roughly 25% of questions. Walli’s formula looks intimidating but becomes mechanical after 10 practice problems. Leibniz rule questions in JEE Main are usually straightforward — just differentiation under the integral sign.

Week 3 — Limit of sum. This topic has a very specific question structure. Once you practice converting 15-20 such sums to integrals, you’ll recognise them instantly in the exam.

Time management in the exam: If a definite integral question takes more than 2.5 minutes, you’re likely missing a property trick. Step back, check the limits for symmetry, check if King’s property applies, check if the integrand is even/odd. Don’t brute-force compute — that’s the trap.

The 90-second rule: In JEE Main, a well-prepared student should solve most definite integral questions in 60-90 seconds. This chapter is a time-saving chapter if you know it well — use those saved minutes on Calculus topics that genuinely need computation.

For JEE Advanced, additionally master: recursion $I_n = f(n) \cdot I_{n-2}$ type problems, comparison theorems (which integral is larger without evaluating), and floor/fractional part integrals over natural number periods.

Common Traps

Trap 1 — Forgetting to check continuity before splitting limits.

Many students write $\int_{-1}^{1} \frac{1}{x}\,dx = 0$ using the odd function property. This is wrong — $\frac{1}{x}$ is discontinuous at $x = 0$ . The integral doesn’t exist. Always check for discontinuities inside the limits before applying properties.

Trap 2 — Wrong application of Walli’s formula for $K$ .

$K = \pi/2$ only when both $m$ and $n$ are even. For $\int_0^{\pi/2} \sin^4 x \cos^3 x\,dx$ , since $n=3$ is odd, $K=1$ . Many students reflexively write $\pi/2$ and lose the mark.

Trap 3 — Limit of sum: missing the $1/n$ factor.

When converting $\lim_{n\to\infty} \frac{1}{n}\sum f(r/n)$ , students sometimes write the integrand without dividing by $n$ . Double-check: the factor $\frac{1}{n}$ must be present (it becomes $dx$ ). Also watch for sums starting at $r=0$ vs $r=1$ — this changes the lower limit from $0$ to $0$ (no change for most cases, but matters when the function has a discontinuity at 0).

Trap 4 — Sign errors with periodic function property.

$\int_a^{a+T} f(x)\,dx = \int_0^T f(x)\,dx$ only when $f$ has period $T$ . Students sometimes use this on $|\sin x|$ (period $\pi$ ) for limits $[0, 3\pi]$ and get $3 \cdot \int_0^{\pi} |\sin x|\,dx = 6$ , which is correct — but they make errors when limits don’t start at 0. Use: $\int_a^{a+nT} f(x)\,dx = n\int_0^T f(x)\,dx$ for any $a$ .

Trap 5 — Leibniz rule with a constant limit.

For $F(x) = \int_2^x f(t)\,dt$ , students sometimes write $F'(x) = f(x) - f(2)$ . The correct answer is just $F'(x) = f(x)$ — the lower limit is a constant, so its derivative is zero (not $f(2)$ ).