Definite Integration JEE Notes 2026 | Properties, King's Property, Wallis Formula, Leibniz Rule, PYQs

1

Definition & Geometric Interpretation

Definite Integration is the process of evaluating the integral of a function between two specified limits. It represents the signed area enclosed between the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$. This is one of the most important and high-scoring chapters in JEE Mathematics — typically 1–2 questions appear every session.

1.1 Definition

The definite integral of a function $f(x)$ from $x = a$ to $x = b$ is denoted by $\int_a^b f(x)\, dx$ and is defined as the limit of a Riemann sum:

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

If the limit exists, $f$ is said to be integrable on $[a, b]$. Every continuous function is integrable.

1.2 Geometric Interpretation

Most Repeated Concept — Geometric Meaning

The definite integral $\int_a^b f(x)\, dx$ represents the signed area between $y = f(x)$ and the x-axis from $x = a$ to $x = b$:

If $f(x) > 0$ on $[a, b]$: area is positive (above x-axis)
If $f(x) < 0$ on $[a, b]$: area is negative (below x-axis)
If $f$ changes sign: integral = (area above) − (area below)
Total geometric area $= \int_a^b |f(x)|\, dx$

\text{Signed Area} = \int_a^b f(x)\, dx, \qquad \text{Total Area} = \int_a^b |f(x)|\, dx

1.3 Notation & Terminology

In $\int_a^b f(x)\, dx$:

$a$: lower limit, $b$: upper limit
$f(x)$: integrand (function being integrated)
$x$: variable of integration (dummy variable)
$\int_a^b f(x)\, dx = \int_a^b f(t)\, dt = \int_a^b f(u)\, du$ (variable is dummy)
The result is a constant (number), unlike indefinite integral which is a function.

1.4 Important Convention

Important Convention

We define (even when $a > b$):

$\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx$
$\int_a^a f(x)\, dx = 0$
$\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx$ (for any $c$)

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Question: The value of $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx$ is:

(a) $\pi/2$

(b) $\pi/4$

(c) $\pi/8$

(d) $0$

Solution:

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

Apply King's Property: replace $x$ by $\frac{\pi}{2} - x$:

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

Adding both:

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

I = \frac{\pi}{4}

\frac{\pi}{4}

Answer: (b) $\pi/4$

2

Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus (FTC) is the central theorem linking differentiation and integration. It has two parts — both are highly relevant for JEE.

2.1 First Fundamental Theorem (FTC 1)

Most Repeated PYQ — FTC 1

If $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\, dt$, then $F$ is differentiable on $(a, b)$ and:

F'(x) = \frac{d}{dx}\int_a^x f(t)\, dt = f(x)

In words: the derivative of the integral function gives back the original function.

2.2 Second Fundamental Theorem (FTC 2 / Newton-Leibniz Formula)

Most Repeated PYQ — Newton-Leibniz Formula

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (i.e., $F'(x) = f(x)$), then:

\int_a^b f(x)\, dx = F(b) - F(a) = \Big[F(x)\Big]_a^b

This is the most useful formula in definite integration — find the antiderivative and evaluate at limits.

2.3 Generalised FTC (Variable Limits)

Most Repeated PYQ — Variable Limit Derivative

If $F(x) = \int_{a(x)}^{b(x)} f(t)\, dt$, where $a(x)$ and $b(x)$ are differentiable, then:

F'(x) = f(b(x))\cdot b'(x) - f(a(x))\cdot a'(x)

Special cases:

$\frac{d}{dx}\int_a^x f(t)\, dt = f(x)$
$\frac{d}{dx}\int_x^b f(t)\, dt = -f(x)$
$\frac{d}{dx}\int_a^b f(t)\, dt = 0$ (constant)

2.4 Proof Sketch of FTC

For FTC 1: Let $F(x) = \int_a^x f(t)\, dt$. Then:

F(x+h) - F(x) = \int_x^{x+h} f(t)\, dt = f(c)\cdot h \text{ (by MVT for integrals)}

\frac{F(x+h)-F(x)}{h} = f(c) \to f(x) \text{ as } h \to 0

Hence $F'(x) = f(x)$.

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Question: If $F(x) = \int_0^x \frac{t \sin t}{1 + \cos^2 t}\, dt$, then $F'\!\left(\frac{\pi}{2}\right)$ equals:

(a) $0$

(b) $\pi/2$

(c) $\pi/4$

(d) $1$

Solution:

By FTC 1: $F'(x) = \frac{x \sin x}{1 + \cos^2 x}$.

At $x = \pi/2$:

F'\!\left(\frac{\pi}{2}\right) = \frac{(\pi/2) \sin(\pi/2)}{1 + \cos^2(\pi/2)} = \frac{(\pi/2)\cdot 1}{1 + 0} = \frac{\pi}{2}

\frac{\pi}{2}

Answer: (b) $\pi/2$

3

Basic Properties of Definite Integrals

These are the foundational properties that must be memorized. They appear in almost every problem in some form.

3.1 Linearity Property

Important — Linearity

\int_a^b [\alpha f(x) + \beta g(x)]\, dx = \alpha \int_a^b f(x)\, dx + \beta \int_a^b g(x)\, dx

where $\alpha, \beta$ are constants.

3.2 Interval Splitting (Additive) Property

Most Repeated PYQ — Interval Splitting

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

This holds for any $c$ (even outside $[a, b]$), as long as $f$ is integrable on the relevant intervals.

3.3 Reversal of Limits

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

And $\int_a^a f(x)\, dx = 0$.

3.4 Comparison Properties

If $f(x) \leq g(x)$ for all $x \in [a, b]$ (where $a < b$), then:

\int_a^b f(x)\, dx \leq \int_a^b g(x)\, dx

Special case: if $m \leq f(x) \leq M$ on $[a, b]$, then:

m(b-a) \leq \int_a^b f(x)\, dx \leq M(b-a)

3.5 Dummy Variable Property

Dummy Variable

The variable in definite integral is a "dummy" — it can be renamed freely:

\int_a^b f(x)\, dx = \int_a^b f(t)\, dt = \int_a^b f(u)\, du = \int_a^b f(\square)\, d\square

3.6 Integral of Constant

\int_a^b c\, dx = c(b - a)

3.7 Absolute Value Inequality

\left|\int_a^b f(x)\, dx\right| \leq \int_a^b |f(x)|\, dx

PYQ — JEE Main 2024 (Apr 5 Shift 1)

Question: Evaluate $\int_0^2 (3x^2 + 2x + 1)\, dx$.

(a) $12$

(b) $10$

(c) $14$

(d) $8$

Solution:

By linearity:

\int_0^2 (3x^2 + 2x + 1)\, dx = 3\int_0^2 x^2\, dx + 2\int_0^2 x\, dx + \int_0^2 1\, dx

= 3\left[\frac{x^3}{3}\right]_0^2 + 2\left[\frac{x^2}{2}\right]_0^2 + \left[x\right]_0^2

\[= (8 - 0) + (4 - 0) + (2 - 0) = 8 + 4 + 2 = 14\]

\[14\]

Answer: (c) $14$

4

King's Property (Reflection Property)

King's Property is the single most frequently asked property in JEE Mains. Almost every year, at least one question is based on this. The property allows us to "reflect" the integration variable about the midpoint of the interval.

4.1 Statement of King's Property

Most Repeated PYQ — King's Property (Crucial!)

For any integrable function $f$ on $[a, b]$:

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

Proof: Substitute $x = a + b - t$, so $dx = -dt$. When $x = a$, $t = b$; when $x = b$, $t = a$.

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

4.2 Special Case: $\int_0^a$ Form

Most Repeated PYQ — $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

When $a = 0, b = a$ (lower limit 0, upper limit $a$):

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

This is the most commonly used form in JEE problems.

4.3 Application Strategy: Adding Two Integrals

Trick: Add $I$ and $I$

If $I = \int_a^b f(x)\, dx$ is hard to evaluate, apply King's Property to get $I = \int_a^b f(a+b-x)\, dx$. Then:

2I = \int_a^b [f(x) + f(a+b-x)]\, dx

Often the sum $f(x) + f(a+b-x)$ simplifies beautifully (e.g., to a constant).

4.4 Common Patterns Where King's Property Helps

$\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx = \frac{\pi}{4}$
$\int_0^{\pi/2} \frac{\sqrt{\tan x}}{1 + \tan x}\, dx = \frac{\pi}{2\sqrt{2}}$
$\int_0^{\pi/2} \log(\tan x)\, dx = 0$
$\int_0^{\pi/2} \log(\sin x)\, dx = -\frac{\pi}{2}\log 2$
$\int_0^a \frac{f(x)}{f(x) + f(a-x)}\, dx = \frac{a}{2}$

4.5 Generalized King's Property

For periodic function with period $T$, or for $\int_0^{nT}$:

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

Also, $\int_a^{a+T} f(x)\, dx = \int_0^T f(x)\, dx$ for any $a$ (periodic function).

PYQ — JEE Main 2024 (Apr 6 Shift 2)

Question: The value of $\int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x}\, dx$ is:

(a) $\frac{1}{2\sqrt{2}}\log(1+\sqrt{2})$

(b) $\frac{1}{\sqrt{2}}\log(1+\sqrt{2})$

(c) $\frac{1}{2}\log(1+\sqrt{2})$

(d) $\sqrt{2}\log(1+\sqrt{2})$

Solution:

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

Apply King's Property ($x \to \frac{\pi}{2} - x$):

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

Adding:

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

Now $\sin x + \cos x = \sqrt{2}\sin(x + \pi/4)$:

2I = \frac{1}{\sqrt{2}}\int_0^{\pi/2} \csc\!\left(x + \frac{\pi}{4}\right) dx

= \frac{1}{\sqrt{2}}\left[\log\tan\!\left(\frac{x}{2} + \frac{\pi}{8}\right)\right]_0^{\pi/2}

Evaluating: $= \frac{1}{\sqrt{2}}\log\!\left(\frac{\tan(3\pi/8)}{\tan(\pi/8)}\right) = \frac{1}{\sqrt{2}}\log\!\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

$= \frac{1}{\sqrt{2}}\log\!\left((\sqrt{2}+1)^2\right) = \frac{2}{\sqrt{2}}\log(\sqrt{2}+1) = \sqrt{2}\log(\sqrt{2}+1)$

So $I = \frac{\sqrt{2}}{2}\log(\sqrt{2}+1) = \frac{1}{\sqrt{2}}\log(1+\sqrt{2})$.

\frac{1}{\sqrt{2}}\log(1+\sqrt{2})

Answer: (b) $\frac{1}{\sqrt{2}}\log(1+\sqrt{2})$

5

Even & Odd Function Properties

The even/odd nature of a function dramatically simplifies integrals over symmetric limits $[-a, a]$. This is a frequently tested concept.

5.1 Definitions

Even function: $f(-x) = f(x)$ for all $x$ (e.g., $x^2, \cos x, |x|, \sin^2 x$)
Odd function: $f(-x) = -f(x)$ for all $x$ (e.g., $x^3, \sin x, \tan x, x\cos x$)

5.2 Property of Even Function

Most Repeated PYQ — Even Function

If $f$ is even:

\int_{-a}^a f(x)\, dx = 2\int_0^a f(x)\, dx

5.3 Property of Odd Function

Most Repeated PYQ — Odd Function

If $f$ is odd:

\int_{-a}^a f(x)\, dx = 0

5.4 General Property (Combined)

Important — General Form

For any function $f$ (even, odd, or neither):

\int_{-a}^a f(x)\, dx = \int_0^a [f(x) + f(-x)]\, dx

This works even when $f$ is neither even nor odd.

5.5 Key Insight

Trick: Decompose into Even + Odd

Any function can be written as:

f(x) = \underbrace{\frac{f(x)+f(-x)}{2}}_{\text{even part}} + \underbrace{\frac{f(x)-f(-x)}{2}}_{\text{odd part}}

Over $[-a, a]$, only the even part contributes.

PYQ — JEE Main 2023 (Jan 31 Shift 1)

Question: The value of $\int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x\, dx$ is:

(a) $0$

(b) $1$

(c) $2$

(d) $\pi/2$

Solution:

Let $f(x) = \sin^3 x \cos^2 x$.

Check: $f(-x) = \sin^3(-x) \cos^2(-x) = -\sin^3 x \cos^2 x = -f(x)$.

So $f$ is an odd function.

The integral is over symmetric limits $[-\pi/2, \pi/2]$:

\int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x\, dx = 0

\[0\]

Answer: (a) $0$

6

Periodic Function Properties

A function $f$ is periodic with period $T > 0$ if $f(x + T) = f(x)$ for all $x$. Examples: $\sin x, \cos x$ (period $2\pi$), $\tan x$ (period $\pi$), $\{x\}$ (fractional part, period 1).

6.1 Main Periodic Property

Most Repeated PYQ — Periodic Integral

If $f$ has period $T$, then for any integer $n$:

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

6.2 Integral Over Any Period-Length Interval

Most Repeated PYQ — Shift Property

If $f$ has period $T$, then for any real $a$:

\int_a^{a+T} f(x)\, dx = \int_0^T f(x)\, dx

The integral over any interval of length $T$ is the same — this is the most useful form.

6.3 General Periodic Property

If $f$ has period $T$ and $a, b$ are arbitrary:

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

For $b - a = nT + r$ where $0 \leq r < T$:

\int_a^b f(x)\, dx = n\int_0^T f(x)\, dx + \int_0^r f(x)\, dx

6.4 Common Periodic Functions

$\sin x, \cos x$: period $2\pi$, $\int_0^{2\pi} \sin x\, dx = 0$, $\int_0^{2\pi} \cos^2 x\, dx = \pi$
$\tan x, \cot x$: period $\pi$
$|\sin x|, |\cos x|$: period $\pi$
$\sin^2 x, \cos^2 x, \sin^4 x, \cos^4 x$: period $\pi$
$\sin(nx), \cos(nx)$: period $2\pi/n$
$\{x\}$ (fractional part): period $1$
$[x]$ (greatest integer): not periodic, but its "derivative-like" structure repeats

6.5 Useful Result

Important Result

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

Since $|\sin x|$ has period $\pi$ and $\int_0^{\pi} |\sin x|\, dx = 2$.

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Question: The value of $\int_0^{100\pi} |\sin x|\, dx$ is:

(a) $100$

(b) $200$

(c) $50$

(d) $100\pi$

Solution:

$|\sin x|$ has period $\pi$ (since $\sin(x+\pi) = -\sin x$, so $|\sin(x+\pi)| = |\sin x|$).

Using $\int_0^{nT} f(x)\, dx = n\int_0^T f(x)\, dx$ with $T = \pi, n = 100$:

\int_0^{100\pi} |\sin x|\, dx = 100\int_0^{\pi} |\sin x|\, dx

Compute $\int_0^{\pi} |\sin x|\, dx$. Since $\sin x \geq 0$ on $[0, \pi]$:

\int_0^{\pi} |\sin x|\, dx = \int_0^{\pi} \sin x\, dx = [-\cos x]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2

So the integral $= 100 \times 2 = 200$.

\[200\]

Answer: (b) $200$

7

Property: $\int_0^{2a} f(x)\, dx$

This is a very powerful property for integrals of the form $\int_0^{2a}$. It connects to King's property and is heavily tested.

7.1 Main Property

Most Repeated PYQ — $\int_0^{2a}$ Property

For any integrable function $f$:

\int_0^{2a} f(x)\, dx = \int_0^a f(x)\, dx + \int_0^a f(2a - x)\, dx

Equivalently:

\int_0^{2a} f(x)\, dx = \int_0^a \big[f(x) + f(2a - x)\big]\, dx

7.2 Special Cases

Most Repeated PYQ — Special Cases

If $f(2a - x) = f(x)$: $\int_0^{2a} f(x)\, dx = 2\int_0^a f(x)\, dx$
If $f(2a - x) = -f(x)$: $\int_0^{2a} f(x)\, dx = 0$

Example: $f(x) = \sin x$ on $[0, 2\pi]$. Here $2a = 2\pi$, $a = \pi$. $f(2\pi - x) = \sin(2\pi - x) = -\sin x = -f(x)$. So $\int_0^{2\pi} \sin x\, dx = 0$.

7.3 Common Functions and Results

$\int_0^{2\pi} \sin x\, dx = 0$ (since $\sin(2\pi - x) = -\sin x$)
$\int_0^{2\pi} \cos x\, dx = 0$ (since $\cos(2\pi - x) = \cos x$, but $\int_0^{\pi} \cos x\, dx + \int_0^{\pi}\cos(2\pi - x)\, dx = 1 + (-1) = 0$)
$\int_0^{\pi} \sin x\, dx = 2$ (since $\sin(\pi - x) = \sin x$, so $\int_0^{\pi} \sin x\, dx = 2\int_0^{\pi/2}\sin x\, dx = 2$)
$\int_0^{\pi} \cos x\, dx = 0$ (since $\cos(\pi - x) = -\cos x$)
$\int_0^{\pi} x f(\sin x)\, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx$

PYQ — JEE Main 2023 (Apr 11 Shift 2)

Question: The value of $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x}\, dx$ is:

(a) $\pi^2/4$

(b) $\pi^2/2$

(c) $\pi/2$

(d) $\pi^2/8$

Solution:

Use the property $\int_0^{\pi} x f(\sin x)\, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx$.

Here $f(\sin x) = \frac{\sin x}{1 + \cos^2 x}$, which is a function of $\sin x$ only (since $\cos^2 x = 1 - \sin^2 x$, so $1 + \cos^2 x = 2 - \sin^2 x$).

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

Let $u = \cos x$, $du = -\sin x\, dx$. When $x = 0$, $u = 1$; when $x = \pi$, $u = -1$:

\int_0^{\pi} \frac{\sin x}{1 + \cos^2 x}\, dx = \int_1^{-1} \frac{-du}{1 + u^2} = \int_{-1}^{1} \frac{du}{1+u^2} = [\tan^{-1} u]_{-1}^{1}

= \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}

So the original integral $= \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}$.

\frac{\pi^2}{4}

Answer: (a) $\pi^2/4$

8

Modulus, GIF & Fractional Part

Integrals involving modulus $|f(x)|$, greatest integer function $[x]$, and fractional part $\{x\}$ require splitting the interval at points where the function changes definition.

8.1 Modulus Integral

Most Repeated PYQ — Modulus

To evaluate $\int_a^b |f(x)|\, dx$:

Find the points where $f(x) = 0$ in $[a, b]$ — these split the interval.
On each sub-interval, $f$ has constant sign — replace $|f|$ with $\pm f$.
Sum the (positive) areas.

Example: $\int_0^{2\pi} |\sin x|\, dx = \int_0^{\pi}\sin x\, dx + \int_{\pi}^{2\pi}(-\sin x)\, dx = 2 + 2 = 4$.

8.2 Greatest Integer Function $[x]$

Most Repeated PYQ — GIF Integrals

$[x]$ = greatest integer $\leq x$. On $[n, n+1)$, $[x] = n$ (constant). So:

\int_0^n [x]\, dx = \sum_{k=0}^{n-1} \int_k^{k+1} [x]\, dx = \sum_{k=0}^{n-1} k \cdot 1 = \frac{n(n-1)}{2}

More generally:

\int_0^n [x]\, dx = \frac{n(n-1)}{2}, \quad n \in \mathbb{Z}^+

8.3 Fractional Part $\{x\} = x - [x]$

Since $\{x\}$ is periodic with period 1, and $\int_0^1 \{x\}\, dx = \int_0^1 x\, dx = \frac{1}{2}$ (as $[x] = 0$ on $[0, 1)$):

\int_0^n \{x\}\, dx = \frac{n}{2}, \quad n \in \mathbb{Z}^+

8.4 Useful Results

$\int_a^b [x]\, dx = \frac{[b]([b]-1)}{2} - \frac{[a]([a]-1)}{2} + [b]\cdot\{b\} - [a]\cdot\{a\} \cdot ...$ (split carefully)
$\int_0^x [t]\, dt = \frac{[x]([x]-1)}{2} + [x]\{x\}$
$\int_0^x \{t\}\, dt = \frac{\{x\}^2}{2} + \frac{[x]}{2}$ ... (verify)

PYQ — JEE Main 2024 (Apr 8 Shift 2)

Question: The value of $\int_0^{5/2} [x]\, dx$ (where $[\cdot]$ is GIF) is:

(a) $3$

(b) $2$

(c) $5/2$

(d) $1$

Solution:

Split at integer points: $[0, 1), [1, 2), [2, 5/2)$.

\int_0^{5/2} [x]\, dx = \int_0^1 0\, dx + \int_1^2 1\, dx + \int_2^{5/2} 2\, dx

= 0\cdot 1 + 1\cdot 1 + 2 \cdot \frac{1}{2} = 0 + 1 + 1 = 2

\[2\]

Answer: (b) $2$

9

Definite Integral as Limit of Sum

The definite integral is defined as a limit of a sum. This concept is frequently tested in JEE Mains.

9.1 Definition (Limit of Sum)

Most Repeated PYQ — Limit of Sum

For $f$ continuous on $[a, b]$, divide $[a, b]$ into $n$ equal parts of width $h = \frac{b-a}{n}$. Then:

\int_a^b f(x)\, dx = \lim_{n \to \infty} h\big[f(a) + f(a+h) + f(a+2h) + \cdots + f(a + (n-1)h)\big]

Equivalently:

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

9.2 Special Case: $\int_0^1$

Most Repeated PYQ — $\int_0^1$ as Limit

With $a = 0, b = 1, h = 1/n$:

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

Or equivalently:

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

9.3 General Form

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

9.4 Strategy for "Limit of Sum" Problems

Strategy Steps

Step 1: Express the sum in the form $\frac{1}{n}\sum f\!\left(\frac{r}{n}\right)$ or $\frac{(b-a)}{n}\sum f\!\left(a + r\cdot\frac{b-a}{n}\right)$.
Step 2: Identify $f(x)$ and the limits $a, b$.
Step 3: Replace $\frac{r}{n} \to x$, $\frac{1}{n} \to dx$, $\sum \to \int$.
Step 4: As $r$ goes from $1$ to $n$, $\frac{r}{n}$ goes from $\frac{1}{n} \to 0$ to $1$. So limits become $0$ to $1$.

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Question: $\lim_{n \to \infty} \frac{1}{n}\left(\frac{1}{n+1} + \frac{2}{n+2} + \cdots + \frac{n}{2n}\right)$ equals:

(a) $\log 2$

(b) $1 - \log 2$

(c) $2\log 2 - 1$

(d) $2 - \log 2$

Solution:

The sum is $S_n = \sum_{r=1}^n \frac{r}{n+r} = \sum_{r=1}^n \frac{r/n}{1 + r/n} \cdot \frac{1}{n} \cdot n$... wait, let's rewrite:

\frac{1}{n}\sum_{r=1}^n \frac{r}{n+r} = \frac{1}{n}\sum_{r=1}^n \frac{r/n}{1 + r/n}

As $n \to \infty$, this approaches $\int_0^1 \frac{x}{1+x}\, dx$.

\int_0^1 \frac{x}{1+x}\, dx = \int_0^1 \left(1 - \frac{1}{1+x}\right) dx = \Big[x - \log(1+x)\Big]_0^1

= (1 - \log 2) - (0 - 0) = 1 - \log 2

1 - \log 2

Answer: (b) $1 - \log 2$

10

Summation of Series (Sigma → Integral)

Converting infinite series to definite integrals is a heavily tested concept. The trick is recognizing the pattern $\frac{1}{n}\sum f\!\left(\frac{r}{n}\right)$.

10.1 Standard Formula

Most Repeated PYQ — Sigma to Integral

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

General:

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

10.2 Different Forms

$\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\, dx$
$\lim_{n \to \infty} \sum_{r=0}^{n-1} \frac{1}{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\, dx$
$\lim_{n \to \infty} \sum_{r=1}^{kn} \frac{1}{n} f\!\left(\frac{r}{n}\right) = \int_0^k f(x)\, dx$
$\lim_{n \to \infty} \sum_{r=1}^n \frac{b-a}{n} f\!\left(a + \frac{r(b-a)}{n}\right) = \int_a^b f(x)\, dx$

10.3 Common Patterns

$\sum_{r=1}^n \frac{n}{n^2 + r^2} = \sum \frac{1}{n}\cdot\frac{1}{1 + (r/n)^2} \to \int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4}$
$\sum_{r=1}^n \frac{r}{n^2 + r^2} = \sum \frac{1}{n}\cdot\frac{r/n}{1 + (r/n)^2} \to \int_0^1 \frac{x\, dx}{1+x^2} = \frac{\log 2}{2}$
$\sum_{r=1}^n \frac{1}{\sqrt{n^2 - r^2}} \to \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}$
$\sum_{r=1}^n \frac{n}{n^2 + r^2} \cdot \frac{1}{n} \to \int_0^1 \frac{dx}{1+x^2}$

10.4 Finding Limits of Sum

Identify $a$ and $b$

In $\sum_{r=p}^{q} f\!\left(\frac{r}{n}\right) \cdot \frac{1}{n}$:

As $n \to \infty$, $\frac{p}{n} \to 0$ and $\frac{q}{n} \to $ (limit value of $q/n$)
If $q = n$: limits are $0$ to $1$
If $q = 2n$: limits are $0$ to $2$
If $q = kn$: limits are $0$ to $k$

PYQ — JEE Main 2023 (Apr 6 Shift 2)

Question: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2 + r^2}$ equals:

(a) $\pi/4$

(b) $\pi/2$

(c) $\pi$

(d) $\log 2$

Solution:

\sum_{r=1}^{n} \frac{n}{n^2 + r^2} = \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + (r/n)^2}

As $n \to \infty$, this becomes:

\int_0^1 \frac{dx}{1 + x^2} = \Big[\tan^{-1} x\Big]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}

\frac{\pi}{4}

Answer: (a) $\pi/4$

11

Wallis' Formula

Wallis' Formula gives the value of $\int_0^{\pi/2} \sin^n x\, dx$ and $\int_0^{\pi/2} \cos^n x\, dx$. This is one of the most asked formulas in JEE Mains.

11.1 Wallis' Formula Statement

Most Repeated PYQ — Wallis' Formula (Crucial!)

$\int_0^{\pi/2} \sin^n x\, dx = \int_0^{\pi/2} \cos^n x\, dx = I_n$, where:

Case 1: $n$ is odd ($n = 2m+1$):

I_n = \frac{2 \cdot 4 \cdot 6 \cdots (n-1)}{1 \cdot 3 \cdot 5 \cdots n} = \frac{(n-1)!!}{n!!}\text{ (with even double factorial on top)}

Case 2: $n$ is even ($n = 2m$):

I_n = \frac{1 \cdot 3 \cdot 5 \cdots (n-1)}{2 \cdot 4 \cdot 6 \cdots n}\cdot \frac{\pi}{2}

11.2 Examples

$\int_0^{\pi/2} \sin x\, dx = 1$
$\int_0^{\pi/2} \sin^2 x\, dx = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}$
$\int_0^{\pi/2} \sin^3 x\, dx = \frac{2}{3}$
$\int_0^{\pi/2} \sin^4 x\, dx = \frac{1\cdot 3}{2\cdot 4}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$
$\int_0^{\pi/2} \sin^5 x\, dx = \frac{2\cdot 4}{1\cdot 3\cdot 5} = \frac{8}{15}$
$\int_0^{\pi/2} \sin^6 x\, dx = \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\cdot\frac{\pi}{2} = \frac{5\pi}{32}$

11.3 Reduction Formula

Important — Reduction Formula

I_n = \frac{n-1}{n} I_{n-2}, \quad \text{with } I_0 = \frac{\pi}{2}, I_1 = 1

This recurrence relation generates all values of $I_n$.

11.4 Product Formula (Mixed Powers)

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

If both $m$ and $n$ are even: $\frac{(m-1)!!(n-1)!!}{(m+n)!!}\cdot\frac{\pi}{2}$
If at least one is odd: $\frac{(m-1)!!(n-1)!!}{(m+n)!!}$

where $k!! = k(k-2)(k-4)\cdots$ (double factorial, with $0!! = 1$, $(-1)!! = 1$).

11.5 Useful Wallis-Type Results

$\int_0^{\pi/2} \sin x \cos x\, dx = \frac{1}{2}$
$\int_0^{\pi/2} \sin^2 x \cos^2 x\, dx = \frac{\pi}{16}$
$\int_0^{\pi/2} \sin^3 x \cos^2 x\, dx = \frac{2}{15}$
$\int_0^{\pi} \sin^n x\, dx = 2\int_0^{\pi/2} \sin^n x\, dx = 2I_n$ (for $n \geq 0$, since $\sin(\pi - x) = \sin x$)
$\int_0^{\pi} \cos^n x\, dx = 0$ for odd $n$; $= 2I_n$ for even $n$

PYQ — JEE Main 2024 (Apr 9 Shift 1)

Question: The value of $\int_0^{\pi/2} \sin^4 x \cos^2 x\, dx$ is:

(a) $\pi/32$

(b) $\pi/16$

(c) $3\pi/32$

(d) $\pi/8$

Solution:

Both $m=4$ and $n=2$ are even. Use the mixed Wallis formula:

\int_0^{\pi/2} \sin^4 x \cos^2 x\, dx = \frac{(4-1)!!(2-1)!!}{(4+2)!!}\cdot\frac{\pi}{2} = \frac{3!! \cdot 1!!}{6!!}\cdot\frac{\pi}{2}

$3!! = 3\cdot 1 = 3$, $1!! = 1$, $6!! = 6\cdot 4\cdot 2 = 48$.

= \frac{3 \cdot 1}{48}\cdot\frac{\pi}{2} = \frac{3\pi}{96} = \frac{\pi}{32}

Alternative: $\sin^4 x \cos^2 x = \sin^4 x(1 - \sin^2 x) = \sin^4 x - \sin^6 x$.

\int_0^{\pi/2} \sin^4 x\, dx - \int_0^{\pi/2} \sin^6 x\, dx = \frac{3\pi}{16} - \frac{5\pi}{32} = \frac{6\pi - 5\pi}{32} = \frac{\pi}{32}

\frac{\pi}{32}

Answer: (a) $\pi/32$

12

Integration by Parts in Definite Integrals

Integration by Parts (IBP) extends directly to definite integrals with explicit boundary terms.

12.1 Formula

Most Repeated PYQ — IBP in Definite

\int_a^b u\, v\, dx = \Big[u \int v\, dx\Big]_a^b - \int_a^b \left(\frac{du}{dx}\right)\left(\int v\, dx\right) dx

Or equivalently:

\int_a^b u\, v\, dx = \Big[u V\Big]_a^b - \int_a^b u' V\, dx, \quad \text{where } V = \int v\, dx

12.2 ILATE Rule

Choose $u$ (the function to differentiate) by ILATE priority:

I – Inverse trigonometric ($\sin^{-1} x, \tan^{-1} x$, etc.)
L – Logarithmic ($\log x, \ln x$)
A – Algebraic ($x^n$, polynomials)
T – Trigonometric ($\sin x, \cos x$, etc.)
E – Exponential ($e^x, a^x$)

The function appearing earlier in ILATE is chosen as $u$.

12.3 Special Result

\int_0^1 x^n e^x\, dx = e - n\int_0^1 x^{n-1} e^x\, dx

A recursive (reduction) formula.

12.4 Useful IBP Results

$\int_0^1 \log x\, dx = -1$ (using $x\log x - x$ evaluated)
$\int_0^1 x \log x\, dx = -\frac{1}{4}$
$\int_0^1 x^n \log x\, dx = -\frac{1}{(n+1)^2}$
$\int_0^{\pi/2} x \sin x\, dx = 1$
$\int_0^{\pi/2} x \cos x\, dx = \frac{\pi}{2} - 1$

PYQ — JEE Main 2022 (July 28 Shift 1)

Question: The value of $\int_0^1 x \log(1+x)\, dx$ is:

(a) $\frac{1}{4}$

(b) $\frac{1}{4}\log 2 - \frac{1}{4}$ ... wait

(c) $\frac{1}{2}\log 2 - \frac{1}{4}$

(d) $\frac{1}{4}\log 2 + \frac{1}{4}$ ...

Solution:

Use IBP with $u = \log(1+x)$, $dv = x\, dx$. So $du = \frac{dx}{1+x}$, $v = \frac{x^2}{2}$.

\int_0^1 x\log(1+x)\, dx = \left[\frac{x^2}{2}\log(1+x)\right]_0^1 - \int_0^1 \frac{x^2}{2(1+x)}\, dx

= \frac{1}{2}\log 2 - \frac{1}{2}\int_0^1 \frac{x^2}{1+x}\, dx

Now $\frac{x^2}{1+x} = x - 1 + \frac{1}{1+x}$ (polynomial division).

\int_0^1 \frac{x^2}{1+x}\, dx = \int_0^1 \left(x - 1 + \frac{1}{1+x}\right) dx = \frac{1}{2} - 1 + \log 2 = -\frac{1}{2} + \log 2

So the integral $= \frac{1}{2}\log 2 - \frac{1}{2}\left(-\frac{1}{2} + \log 2\right) = \frac{1}{2}\log 2 + \frac{1}{4} - \frac{1}{2}\log 2 = \frac{1}{4}$.

Wait, recompute: $\frac{1}{2}\log 2 - \frac{1}{2}(-1/2 + \log 2) = \frac{1}{2}\log 2 + \frac{1}{4} - \frac{1}{2}\log 2 = \frac{1}{4}$.

\frac{1}{4}

Answer: (a) $\frac{1}{4}$

13

Substitution Method in Definite Integrals

Substitution in definite integrals differs from indefinite in that limits must also be changed (or converted back at the end).

13.1 Method

Most Repeated PYQ — Substitution with Changed Limits

To evaluate $\int_a^b f(g(x))\, g'(x)\, dx$, substitute $t = g(x)$, $dt = g'(x)\, dx$. When $x = a, t = g(a)$; when $x = b, t = g(b)$:

\int_a^b f(g(x))\, g'(x)\, dx = \int_{g(a)}^{g(b)} f(t)\, dt

Key: When you change the variable, change the limits too! No need to revert back.

13.2 Common Substitutions

$\sqrt{a^2 - x^2}$: use $x = a\sin\theta$
$\sqrt{a^2 + x^2}$: use $x = a\tan\theta$
$\sqrt{x^2 - a^2}$: use $x = a\sec\theta$
$\frac{1}{a + b\cos x}$ or $\frac{1}{a + b\sin x}$: use Weierstrass $t = \tan(x/2)$
For $\int_0^{\pi/2} f(\sin x, \cos x)\, dx$: try $x \to \pi/2 - x$ (King's)
For $\int_0^a f(x)\, dx$ with $f$ containing $\sqrt{ax - x^2}$: use $x = a\sin^2\theta$

13.3 Symmetric Substitution Trick

Trick

For $\int_0^a f(x)\, dx$, if $f$ involves $x$ and $a - x$ symmetrically, substitution $x \to a - x$ (King's) often simplifies.

PYQ — JEE Main 2023 (Apr 10 Shift 1)

Question: Evaluate $\int_0^1 \frac{x^4}{\sqrt{1 - x^2}}\, dx$.

(a) $3\pi/16$

(b) $\pi/4$

(c) $3\pi/8$

(d) $\pi/16$

Solution:

Substitute $x = \sin\theta$, $dx = \cos\theta\, d\theta$.

When $x = 0, \theta = 0$; when $x = 1, \theta = \pi/2$.

\int_0^1 \frac{x^4}{\sqrt{1-x^2}}\, dx = \int_0^{\pi/2} \frac{\sin^4\theta}{\cos\theta}\cos\theta\, d\theta = \int_0^{\pi/2} \sin^4\theta\, d\theta

By Wallis' formula (n = 4, even):

\int_0^{\pi/2} \sin^4\theta\, d\theta = \frac{1\cdot 3}{2\cdot 4}\cdot\frac{\pi}{2} = \frac{3\pi}{16}

\frac{3\pi}{16}

Answer: (a) $3\pi/16$

14

Leibniz Rule (Differentiation under Integral Sign)

Leibniz Rule (also called Newton-Leibniz formula or Differentiation under the integral sign) is heavily tested in JEE Advanced and frequently in JEE Mains.

14.1 Basic Leibniz Rule

Most Repeated PYQ — Leibniz Rule (Crucial!)

If $F(x) = \int_{a(x)}^{b(x)} f(t)\, dt$, where $a(x), b(x)$ are differentiable and $f$ is continuous, then:

F'(x) = f(b(x))\cdot b'(x) - f(a(x))\cdot a'(x)

14.2 Special Cases

$\frac{d}{dx}\int_a^x f(t)\, dt = f(x)$
$\frac{d}{dx}\int_x^b f(t)\, dt = -f(x)$
$\frac{d}{dx}\int_a^b f(t)\, dt = 0$ (constant)
$\frac{d}{dx}\int_{-x}^x f(t)\, dt = f(x) + f(-x)$

14.3 Extended Leibniz Rule (Integrand Depends on $x$)

Important — Extended Form

If $F(x) = \int_{a(x)}^{b(x)} f(x, t)\, dt$ (integrand depends on $x$ too):

F'(x) = f(x, b(x))\cdot b'(x) - f(x, a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(x, t)\, dt

14.4 Strategy for Leibniz Problems

Problem-Solving Approach

Identify $a(x)$ and $b(x)$ — the lower and upper limits.
Compute $a'(x)$ and $b'(x)$.
Substitute $b(x)$ into $f$ to get $f(b(x))$.
Substitute $a(x)$ into $f$ to get $f(a(x))$.
Apply: $F'(x) = f(b)\cdot b' - f(a)\cdot a'$.

14.5 Common Application: Functional Equations

If $f(x) = \int_0^x g(t)\, dt$ and we know $g$, we can find $f$ by integration. Conversely, if $f$ is given via an integral equation, differentiate to find $g$.

PYQ — JEE Main 2024 (Jan 29 Shift 2)

Question: Let $f(x) = \int_0^{x^2} \frac{t}{1 + t^2}\, dt$. Then $f'(1)$ equals:

(a) $\frac{1}{2}$

(b) $1$

(c) $\frac{2}{3}$

(d) $\frac{1}{4}$

Solution:

Here $a(x) = 0$ (constant) and $b(x) = x^2$.

By Leibniz: $f'(x) = \frac{b(x)}{1 + b(x)^2}\cdot b'(x) - 0 = \frac{x^2}{1 + x^4}\cdot 2x = \frac{2x^3}{1+x^4}$.

At $x = 1$:

f'(1) = \frac{2\cdot 1}{1 + 1} = 1

\[1\]

Answer: (b) $1$

15

Reduction Formulas

Reduction formulas express an integral of higher power in terms of an integral of lower power, allowing recursive evaluation.

15.1 $\int_0^{\pi/2} \sin^n x\, dx$ (Wallis Reduction)

Most Repeated PYQ — Reduction Formula

Let $I_n = \int_0^{\pi/2} \sin^n x\, dx$. Then:

I_n = \frac{n-1}{n} I_{n-2}, \quad I_0 = \frac{\pi}{2}, \quad I_1 = 1

15.2 $\int_0^{\pi/2} \cos^n x\, dx$

Same recurrence (since $\int_0^{\pi/2}\cos^n x\, dx = \int_0^{\pi/2}\sin^n x\, dx$ by King's):

J_n = \frac{n-1}{n} J_{n-2}, \quad J_0 = \frac{\pi}{2}, \quad J_1 = 1

15.3 $\int_0^{\pi/2} \sin^m x \cos^n x\, dx$

Important — Mixed Reduction

I_{m,n} = \frac{m-1}{m+n} I_{m-2, n} = \frac{n-1}{m+n} I_{m, n-2}

15.4 Other Reduction Formulas

$\int \tan^n x\, dx = \frac{\tan^{n-1} x}{n-1} - \int \tan^{n-2} x\, dx$
$\int \sec^n x\, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2} x\, dx$
$\int \cot^n x\, dx = -\frac{\cot^{n-1} x}{n-1} - \int \cot^{n-2} x\, dx$
$\int \csc^n x\, dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2} x\, dx$
$\int_0^1 x^n e^x\, dx = e - n\int_0^1 x^{n-1} e^x\, dx$
$\int_0^1 (\log x)^n\, dx = (-1)^n n!$

15.5 Derivation of Wallis Reduction

Quick Derivation

$I_n = \int_0^{\pi/2} \sin^n x\, dx = \int_0^{\pi/2} \sin^{n-1} x \cdot \sin x\, dx$.

Apply IBP with $u = \sin^{n-1}x$, $dv = \sin x\, dx$:

I_n = \Big[-\sin^{n-1} x \cos x\Big]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2} x \cos^2 x\, dx

Boundary term = 0. And $\cos^2 x = 1 - \sin^2 x$:

I_n = (n-1)\int_0^{\pi/2} \sin^{n-2} x\, dx - (n-1)\int_0^{\pi/2}\sin^n x\, dx = (n-1)I_{n-2} - (n-1)I_n

So $I_n(1 + n - 1) = (n-1)I_{n-2}$, giving $I_n = \frac{n-1}{n}I_{n-2}$.

PYQ — JEE Main 2022 (June 27 Shift 2)

Question: Using reduction formula, $\int_0^{\pi/2} \sin^7 x\, dx$ equals:

(a) $\frac{16}{35}$

(b) $\frac{8}{35}$

(c) $\frac{16}{105}$

(d) $\frac{32}{35}$

Solution:

For $n = 7$ (odd), Wallis' formula:

I_7 = \frac{2\cdot 4\cdot 6}{1\cdot 3\cdot 5\cdot 7} = \frac{48}{105} = \frac{16}{35}

Using reduction: $I_7 = \frac{6}{7}I_5 = \frac{6}{7}\cdot\frac{4}{5}I_3 = \frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}I_1 = \frac{6\cdot 4\cdot 2}{7\cdot 5\cdot 3}\cdot 1 = \frac{48}{105} = \frac{16}{35}$.

\frac{16}{35}

Answer: (a) $\frac{16}{35}$

16

Special Definite Integrals (Log sin, Log cos)

These special integrals involving logarithms of trigonometric functions are high-yield — almost guaranteed to be tested.

16.1 $\int_0^{\pi/2} \log(\sin x)\, dx$

Most Repeated PYQ — Log sin x (Crucial!)

\int_0^{\pi/2} \log(\sin x)\, dx = -\frac{\pi}{2}\log 2

Proof: Let $I = \int_0^{\pi/2} \log(\sin x)\, dx$. By King's property, $I = \int_0^{\pi/2}\log(\cos x)\, dx$. Adding:

2I = \int_0^{\pi/2} \log(\sin x \cos x)\, dx = \int_0^{\pi/2} \log\!\left(\frac{\sin 2x}{2}\right) dx = \int_0^{\pi/2}\log(\sin 2x)\, dx - \frac{\pi}{2}\log 2

In the first integral, substitute $u = 2x$, $du = 2\, dx$. When $x = 0, u = 0$; when $x = \pi/2, u = \pi$:

\int_0^{\pi/2}\log(\sin 2x)\, dx = \frac{1}{2}\int_0^{\pi}\log(\sin u)\, du = \frac{1}{2}\cdot 2\int_0^{\pi/2}\log(\sin u)\, du = I

So $2I = I - \frac{\pi}{2}\log 2$, giving $I = -\frac{\pi}{2}\log 2$.

16.2 $\int_0^{\pi/2} \log(\cos x)\, dx$

\int_0^{\pi/2} \log(\cos x)\, dx = -\frac{\pi}{2}\log 2

(Same as $\log\sin$ by King's property: $\cos x = \sin(\pi/2 - x)$.)

16.3 $\int_0^{\pi/2} \log(\tan x)\, dx$

Most Repeated PYQ — Log tan x

\int_0^{\pi/2} \log(\tan x)\, dx = 0

Proof: $\log(\tan x) = \log(\sin x) - \log(\cos x)$. Since both integrals are equal, the difference is 0.

16.4 Other Log-Trig Integrals

$\int_0^{\pi} \log(\sin x)\, dx = -\pi \log 2$
$\int_0^{\pi/2} \log(\sec x)\, dx = \frac{\pi}{2}\log 2$
$\int_0^{\pi/2} \log(\csc x)\, dx = \frac{\pi}{2}\log 2$
$\int_0^{\pi/2} \log(1 + \cos x)\, dx = -\frac{\pi}{2}\log 2 + 2G$ (Catalan's constant, advanced)
$\int_0^{\pi/2} x\log(\sin x)\, dx = \frac{\pi^2}{8}\log 2 - \frac{7}{16}\zeta(3)$ (advanced)

16.5 Squared Log Integrals

$\int_0^{\pi/2} \log^2(\sin x)\, dx = \frac{\pi}{2}\log^2 2 + \frac{\pi^3}{24}$
$\int_0^1 \frac{\log x}{1 + x}\, dx = -\frac{\pi^2}{12}$
$\int_0^1 \frac{\log x}{1 - x}\, dx = -\frac{\pi^2}{6}$

PYQ — JEE Main 2023 (Apr 12 Shift 1)

Question: The value of $\int_0^{\pi/2} \log(\tan x)\, dx$ is:

(a) $0$

(b) $-\frac{\pi}{2}\log 2$

(c) $\frac{\pi}{2}\log 2$

(d) $\pi\log 2$

Solution:

$\log(\tan x) = \log(\sin x) - \log(\cos x)$.

\int_0^{\pi/2} \log(\tan x)\, dx = \int_0^{\pi/2}\log(\sin x)\, dx - \int_0^{\pi/2}\log(\cos x)\, dx

By King's property, both integrals are equal (to $-\frac{\pi}{2}\log 2$).

= -\frac{\pi}{2}\log 2 - \left(-\frac{\pi}{2}\log 2\right) = 0

\[0\]

Answer: (a) $0$

17

Property: $\int_0^{\pi} x\, f(\sin x)\, dx$

This is a special property for integrals of the form $\int_0^{\pi} x f(\sin x)\, dx$. Very frequently tested in JEE.

17.1 Main Property

Most Repeated PYQ — π f(sin x) Property

If $f$ is any function:

\int_0^{\pi} x\, f(\sin x)\, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx

Proof: Let $I = \int_0^{\pi} x\, f(\sin x)\, dx$. By King's property ($x \to \pi - x$):

I = \int_0^{\pi} (\pi - x) f(\sin(\pi - x))\, dx = \int_0^{\pi}(\pi - x) f(\sin x)\, dx

I = \pi\int_0^{\pi} f(\sin x)\, dx - I

2I = \pi\int_0^{\pi} f(\sin x)\, dx \implies I = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx

17.2 Generalization

More generally:

\int_0^{2a} x\, f(x)\, dx = a\int_0^{2a} f(x)\, dx \quad \text{if } f(2a - x) = f(x)

17.3 Common Examples

$\int_0^{\pi} x \sin x\, dx = \frac{\pi}{2}\int_0^{\pi}\sin x\, dx = \frac{\pi}{2}\cdot 2 = \pi$
$\int_0^{\pi} x \sin^3 x\, dx = \frac{\pi}{2}\int_0^{\pi}\sin^3 x\, dx = \frac{\pi}{2}\cdot\frac{4}{3} = \frac{2\pi}{3}$
$\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x}\, dx = \frac{\pi}{2}\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\, dx = \frac{\pi^2}{4}$

PYQ — JEE Main 2022 (July 27 Shift 1)

Question: $\int_0^{\pi} x \sin^4 x\, dx$ equals:

(a) $\frac{\pi^2}{4}$

(b) $\frac{3\pi^2}{16}$

(c) $\frac{\pi^2}{8}$

(d) $\frac{3\pi^2}{8}$

Solution:

By the property $\int_0^{\pi} x f(\sin x)\, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx$ with $f(\sin x) = \sin^4 x$:

\int_0^{\pi} x \sin^4 x\, dx = \frac{\pi}{2}\int_0^{\pi} \sin^4 x\, dx

Since $\sin^4(\pi - x) = \sin^4 x$, $\int_0^{\pi}\sin^4 x\, dx = 2\int_0^{\pi/2}\sin^4 x\, dx$.

= 2\cdot\frac{3\pi}{16} = \frac{3\pi}{8}

So $\int_0^{\pi} x \sin^4 x\, dx = \frac{\pi}{2}\cdot\frac{3\pi}{8} = \frac{3\pi^2}{16}$.

\frac{3\pi^2}{16}

Answer: (b) $\frac{3\pi^2}{16}$

18

Inequalities in Definite Integrals

Inequalities help bound integrals without computing them exactly. Useful for JEE Advanced problems.

18.1 Basic Bounds

Important — Bounds

If $m \leq f(x) \leq M$ on $[a, b]$:

m(b - a) \leq \int_a^b f(x)\, dx \leq M(b - a)

18.2 Comparison Property

If $f(x) \leq g(x)$ on $[a, b]$ (with $a < b$):

\int_a^b f(x)\, dx \leq \int_a^b g(x)\, dx

18.3 Absolute Value Inequality

\left|\int_a^b f(x)\, dx\right| \leq \int_a^b |f(x)|\, dx

18.4 Cauchy-Schwarz Inequality

Most Repeated PYQ — Cauchy-Schwarz

\left(\int_a^b f(x) g(x)\, dx\right)^2 \leq \left(\int_a^b f^2(x)\, dx\right)\left(\int_a^b g^2(x)\, dx\right)

18.5 Useful Trig Inequalities

$\sin x \leq x \leq \tan x$ for $0 < x < \pi/2$
$\cos x \leq \frac{\sin x}{x} \leq 1$ for $0 < x < \pi/2$
$\frac{2}{\pi} \leq \frac{\sin x}{x} \leq 1$ for $0 < x \leq \pi/2$ (Jordan's inequality)

PYQ — JEE Main 2021 (Aug 27 Shift 1)

Question: The inequality $\frac{2}{\pi} \leq \frac{\sin x}{x} \leq 1$ for $x \in (0, \pi/2]$ implies:

(a) $\int_0^{\pi/2} \frac{\sin x}{x}\, dx < 1$

(b) $\int_0^{\pi/2} \frac{\sin x}{x}\, dx \geq 1$

(c) $\int_0^{\pi/2} \frac{\sin x}{x}\, dx > \pi/2$

(d) $\int_0^{\pi/2} \frac{\sin x}{x}\, dx \leq \pi/2$

Solution:

Since $\frac{\sin x}{x} \leq 1$ on $(0, \pi/2]$:

\int_0^{\pi/2} \frac{\sin x}{x}\, dx \leq \int_0^{\pi/2} 1\, dx = \frac{\pi}{2}

And since $\frac{\sin x}{x} \geq \frac{2}{\pi}$:

\int_0^{\pi/2} \frac{\sin x}{x}\, dx \geq \frac{2}{\pi}\cdot\frac{\pi}{2} = 1

So $1 \leq \int_0^{\pi/2} \frac{\sin x}{x}\, dx \leq \frac{\pi}{2}$.

\text{Both (b) and (d) hold}

Answer: (b) and (d) both valid; usually (b) is the intended answer

19

Average Value of Function

The average (mean) value of a function over an interval is a useful concept linking integrals to averages.

19.1 Definition

Important — Mean Value

The average value of $f$ on $[a, b]$ is:

f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\, dx

19.2 Mean Value Theorem for Integrals

Most Repeated PYQ — MVT for Integrals

If $f$ is continuous on $[a, b]$, then there exists $c \in (a, b)$ such that:

f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx

i.e., $f$ attains its average value at some point in the interval.

19.3 RMS Value

The Root Mean Square value of $f$ on $[a, b]$ is:

f_{\text{rms}} = \sqrt{\frac{1}{b-a}\int_a^b f^2(x)\, dx}

19.4 Examples

Average of $\sin x$ on $[0, \pi]$: $\frac{1}{\pi}\int_0^{\pi}\sin x\, dx = \frac{2}{\pi}$
Average of $\sin^2 x$ on $[0, 2\pi]$: $\frac{1}{2\pi}\int_0^{2\pi}\sin^2 x\, dx = \frac{1}{2}$
Average of $\cos^2 x$ on $[0, 2\pi]$: $\frac{1}{2}$
RMS of $\sin x$ on $[0, 2\pi]$: $\sqrt{\frac{1}{2\pi}\int_0^{2\pi}\sin^2 x\, dx} = \frac{1}{\sqrt{2}}$

PYQ — JEE Main 2020 (Sep 4 Shift 2)

Question: The average value of $f(x) = x^2$ on $[0, 2]$ is:

(a) $2/3$

(b) $4/3$

(c) $8/3$

(d) $2$

Solution:

f_{\text{avg}} = \frac{1}{2-0}\int_0^2 x^2\, dx = \frac{1}{2}\left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2}\cdot\frac{8}{3} = \frac{4}{3}

\frac{4}{3}

Answer: (b) $4/3$

20

Important Results & Formula Summary

A consolidated list of all important results for quick revision.

Most Repeated — All Formulas

1. FTC: $\int_a^b f(x)\, dx = F(b) - F(a)$

2. King's Property: $\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx$

3. Even: $\int_{-a}^a f(x)\, dx = 2\int_0^a f(x)\, dx$

4. Odd: $\int_{-a}^a f(x)\, dx = 0$

5. Periodic: $\int_0^{nT} f(x)\, dx = n\int_0^T f(x)\, dx$

6. Any period: $\int_a^{a+T} f(x)\, dx = \int_0^T f(x)\, dx$

7. $\int_0^{2a}$: $= \int_0^a [f(x) + f(2a-x)]\, dx$

8. $\int_0^{\pi} x f(\sin x)$: $= \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx$

9. Wallis (odd $n$): $\int_0^{\pi/2}\sin^n x\, dx = \frac{2\cdot4\cdots(n-1)}{1\cdot3\cdots n}$

10. Wallis (even $n$): $\int_0^{\pi/2}\sin^n x\, dx = \frac{1\cdot3\cdots(n-1)}{2\cdot4\cdots n}\cdot\frac{\pi}{2}$

11. Reduction: $I_n = \frac{n-1}{n} I_{n-2}$

12. Leibniz: $\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)\, dt = f(b)b' - f(a)a'$

13. Log sin: $\int_0^{\pi/2}\log(\sin x)\, dx = -\frac{\pi}{2}\log 2$

14. Log cos: $\int_0^{\pi/2}\log(\cos x)\, dx = -\frac{\pi}{2}\log 2$

15. Log tan: $\int_0^{\pi/2}\log(\tan x)\, dx = 0$

16. Limit of sum: $\lim_{n\to\infty}\frac{1}{n}\sum f(r/n) = \int_0^1 f(x)\, dx$

17. MVT: $\exists c \in (a,b): f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx$

18. Cauchy-Schwarz: $(\int fg)^2 \leq (\int f^2)(\int g^2)$

19. GIF: $\int_0^n [x]\, dx = \frac{n(n-1)}{2}$

20. |sin|: $\int_0^{n\pi}|\sin x|\, dx = 2n$

PYQ — JEE Main 2024 (Apr 9 Shift 2)

Question: Evaluate $\int_{-\pi/2}^{\pi/2} \log\!\left(\frac{2 - \sin x}{2 + \sin x}\right) dx$.

(a) $0$

(b) $\log 2$

(c) $2\log 2$

(d) $\pi\log 2$

Solution:

Let $f(x) = \log\!\left(\frac{2-\sin x}{2+\sin x}\right) = \log(2-\sin x) - \log(2+\sin x)$.

Check even/odd: $f(-x) = \log(2-\sin(-x)) - \log(2+\sin(-x)) = \log(2+\sin x) - \log(2-\sin x) = -f(x)$.

So $f$ is odd.

Integral over symmetric limits $[-\pi/2, \pi/2]$:

\int_{-\pi/2}^{\pi/2} f(x)\, dx = 0

\[0\]

Answer: (a) $0$

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ King's Property: 22%
✓ Wallis Formula: 15%
✓ Limit of Sum / Sigma: 12%
✓ Periodic Functions: 11%
✓ Leibniz Rule: 10%
✓ Even/Odd Functions: 9%
✓ Special Integrals (log sin, log cos): 8%
✓ Modulus / GIF: 7%
✓ $\int_0^{\pi} x f(\sin x)$: 6%

JEE Advanced (Last 10 Years)

✓ Leibniz Rule + Functional Eq: 25%
✓ King's Property (combined): 18%
✓ Reduction Formulas: 15%
✓ Modulus + GIF combined: 12%
✓ Periodic + multi-concept: 12%
✓ Special Integrals: 8%
✓ Inequalities (Cauchy-Schwarz): 5%
✓ Integration by Parts: 5%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 4-12 marks (1-3 questions)

Difficulty Level: Easy to Moderate

Time Required: 6-7 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx$.
Find $\int_0^{\pi/2} \sin^4 x\, dx$ using Wallis.
Evaluate $\int_{-1}^1 x^3 \cos x\, dx$.
Find $\int_0^{2\pi} |\sin x|\, dx$.
Compute $\int_0^3 [x]\, dx$ where $[\cdot]$ is GIF.
Evaluate $\int_0^1 \frac{dx}{1+x^2}$.

Level 2: Intermediate

Evaluate $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x}\, dx$.
Find $\lim_{n\to\infty}\sum_{r=1}^n \frac{n}{n^2+r^2}$.
If $f(x) = \int_0^{x^2} \frac{dt}{1+t}$, find $f'(1)$.
Evaluate $\int_0^{\pi/2} \log(\tan x)\, dx$.
Compute $\int_0^{\pi/2} \sin^3 x \cos^2 x\, dx$.
Find $\int_0^{100\pi} |\cos x|\, dx$.

Level 3: Advanced (JEE Advanced)

Prove $\int_0^{\pi/2} \log(\sin x)\, dx = -\frac{\pi}{2}\log 2$ using King's property.
If $f(x) + f(y) = f(x+y)$ for all $x, y$ and $\int_0^2 f(x)\, dx = 3$, find $f(3)$.
Evaluate $\int_0^{\pi/4} \log(1+\tan x)\, dx$.
Find $\int_0^1 \frac{\log x}{\sqrt{1-x^2}}\, dx$.
Prove that $\int_0^{\pi/2} \frac{dx}{1+\tan^a x} = \frac{\pi}{4}$ for any real $a$.
Show $\int_0^{\pi} x f(\sin x)\, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\, dx$.

Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is King's Property in definite integration?

King's Property states that ∫[a to b] f(x) dx = ∫[a to b] f(a+b-x) dx. This is one of the most important and frequently asked properties in JEE Mains. A special case is ∫[0 to a] f(x) dx = ∫[0 to a] f(a-x) dx.

What is Wallis' formula?

Wallis' formula gives ∫[0 to π/2] sin^n x dx = ∫[0 to π/2] cos^n x dx. For n odd: result = (2·4·6...(n-1))/(1·3·5...n). For n even: result = (1·3·5...(n-1))/(2·4·6...n) · (π/2).

What is Leibniz rule?

Leibniz rule: d/dx ∫[a(x) to b(x)] f(t) dt = f(b(x))·b'(x) - f(a(x))·a'(x). If limits are constants, the derivative is 0. This is highly tested in JEE Advanced.

What is the value of ∫[0 to π/2] log(sin x) dx?

∫[0 to π/2] log(sin x) dx = ∫[0 to π/2] log(cos x) dx = -(π/2) log 2. Also, ∫[0 to π/2] log(tan x) dx = 0 (by King's property).

How to convert a summation into a definite integral?

Use lim(n→∞) (1/n) Σ f(r/n) = ∫[0 to 1] f(x) dx. More generally, lim(n→∞) Σ f(a + r·(b-a)/n) · (b-a)/n = ∫[a to b] f(x) dx.

What is the property of periodic functions?

If f has period T, then ∫[0 to nT] f(x) dx = n ∫[0 to T] f(x) dx. Also ∫[a to a+T] f(x) dx = ∫[0 to T] f(x) dx for any a, i.e., integral over any interval of length T is the same.

How many questions come from Definite Integration in JEE Mains?

On average 1-2 questions (4-8 marks) appear from Definite Integration in every JEE Main session. Most repeated topics are: King's property, periodic functions, Wallis formula, limit of sum, and Leibniz rule.

What is the Fundamental Theorem of Calculus?

FTC Part 1: If F(x) = ∫[a to x] f(t) dt, then F'(x) = f(x). FTC Part 2: If F is an antiderivative of f, then ∫[a to b] f(x) dx = F(b) - F(a). This connects differentiation and integration.

What is ∫[-a to a] f(x) dx for even and odd functions?

If f is even (f(-x) = f(x)): ∫[-a to a] f(x) dx = 2∫[0 to a] f(x) dx. If f is odd (f(-x) = -f(x)): ∫[-a to a] f(x) dx = 0.

What is the property ∫[0 to 2a] f(x) dx?

∫[0 to 2a] f(x) dx = ∫[0 to a] f(x) dx + ∫[0 to a] f(2a-x) dx. If f(2a-x) = -f(x), result is 0. If f(2a-x) = f(x), result is 2∫[0 to a] f(x) dx.

Definite Integration notes, properties & PYQs for JEE 2026

📑 Table of Contents

Definition & Geometric Interpretation

1.1 Definition

1.2 Geometric Interpretation

🔥 Most Repeated Concept — Geometric Meaning

1.3 Notation & Terminology

1.4 Important Convention

⭐ Important Convention

📝 PYQ — JEE Main 2024 (Jan 30 Shift 2)

Fundamental Theorem of Calculus (FTC)

2.1 First Fundamental Theorem (FTC 1)

🔥 Most Repeated PYQ — FTC 1

2.2 Second Fundamental Theorem (FTC 2 / Newton-Leibniz Formula)

🔥 Most Repeated PYQ — Newton-Leibniz Formula

2.3 Generalised FTC (Variable Limits)

🔥 Most Repeated PYQ — Variable Limit Derivative

2.4 Proof Sketch of FTC

📝 PYQ — JEE Main 2023 (Apr 13 Shift 1)

Basic Properties of Definite Integrals

3.1 Linearity Property

⭐ Important — Linearity

3.2 Interval Splitting (Additive) Property

🔥 Most Repeated PYQ — Interval Splitting

3.3 Reversal of Limits

3.4 Comparison Properties

3.5 Dummy Variable Property

💡 Dummy Variable

3.6 Integral of Constant

3.7 Absolute Value Inequality

📝 PYQ — JEE Main 2024 (Apr 5 Shift 1)

King's Property (Reflection Property)

4.1 Statement of King's Property

🔥 Most Repeated PYQ — King's Property (Crucial!)

4.2 Special Case: $\int_0^a$ Form

🔥 Most Repeated PYQ — $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

4.3 Application Strategy: Adding Two Integrals

💡 Trick: Add $I$ and $I$

4.4 Common Patterns Where King's Property Helps

4.5 Generalized King's Property

📝 PYQ — JEE Main 2024 (Apr 6 Shift 2)

Even & Odd Function Properties

5.1 Definitions

5.2 Property of Even Function

🔥 Most Repeated PYQ — Even Function

5.3 Property of Odd Function

🔥 Most Repeated PYQ — Odd Function

5.4 General Property (Combined)

⭐ Important — General Form

5.5 Key Insight

💡 Trick: Decompose into Even + Odd

📝 PYQ — JEE Main 2023 (Jan 31 Shift 1)

Periodic Function Properties

6.1 Main Periodic Property

🔥 Most Repeated PYQ — Periodic Integral

6.2 Integral Over Any Period-Length Interval

🔥 Most Repeated PYQ — Shift Property

6.3 General Periodic Property

6.4 Common Periodic Functions

6.5 Useful Result

⭐ Important Result

📝 PYQ — JEE Main 2024 (Jan 31 Shift 1)

Property: $\int_0^{2a} f(x)\, dx$

7.1 Main Property

🔥 Most Repeated PYQ — $\int_0^{2a}$ Property

7.2 Special Cases

🔥 Most Repeated PYQ — Special Cases

7.3 Common Functions and Results

📝 PYQ — JEE Main 2023 (Apr 11 Shift 2)

Modulus, GIF & Fractional Part

8.1 Modulus Integral

🔥 Most Repeated PYQ — Modulus

8.2 Greatest Integer Function $[x]$

🔥 Most Repeated PYQ — GIF Integrals

8.3 Fractional Part $\{x\} = x - [x]$

8.4 Useful Results

📝 PYQ — JEE Main 2024 (Apr 8 Shift 2)

Definite Integral as Limit of Sum

9.1 Definition (Limit of Sum)

🔥 Most Repeated PYQ — Limit of Sum

Table of Contents

Most Repeated Concept — Geometric Meaning

Important Convention

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Most Repeated PYQ — FTC 1

Most Repeated PYQ — Newton-Leibniz Formula

Most Repeated PYQ — Variable Limit Derivative

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Important — Linearity

Most Repeated PYQ — Interval Splitting

Dummy Variable

PYQ — JEE Main 2024 (Apr 5 Shift 1)

Most Repeated PYQ — King's Property (Crucial!)

Most Repeated PYQ — $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

Trick: Add $I$ and $I$

PYQ — JEE Main 2024 (Apr 6 Shift 2)

Most Repeated PYQ — Even Function

Most Repeated PYQ — Odd Function

Important — General Form

Trick: Decompose into Even + Odd

PYQ — JEE Main 2023 (Jan 31 Shift 1)

Most Repeated PYQ — Periodic Integral

Most Repeated PYQ — Shift Property

Important Result

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Most Repeated PYQ — $\int_0^{2a}$ Property

Most Repeated PYQ — Special Cases

PYQ — JEE Main 2023 (Apr 11 Shift 2)

Most Repeated PYQ — Modulus

Most Repeated PYQ — GIF Integrals

PYQ — JEE Main 2024 (Apr 8 Shift 2)

Most Repeated PYQ — Limit of Sum

Most Repeated PYQ — $\int_0^1$ as Limit

Strategy Steps

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Most Repeated PYQ — Sigma to Integral

Identify $a$ and $b$

PYQ — JEE Main 2023 (Apr 6 Shift 2)

Most Repeated PYQ — Wallis' Formula (Crucial!)

Important — Reduction Formula

PYQ — JEE Main 2024 (Apr 9 Shift 1)

Most Repeated PYQ — IBP in Definite

PYQ — JEE Main 2022 (July 28 Shift 1)

Most Repeated PYQ — Substitution with Changed Limits

Trick

PYQ — JEE Main 2023 (Apr 10 Shift 1)

Most Repeated PYQ — Leibniz Rule (Crucial!)

Important — Extended Form

Problem-Solving Approach

PYQ — JEE Main 2024 (Jan 29 Shift 2)

Most Repeated PYQ — Reduction Formula

Important — Mixed Reduction

Quick Derivation

PYQ — JEE Main 2022 (June 27 Shift 2)

Most Repeated PYQ — Log sin x (Crucial!)

Most Repeated PYQ — Log tan x

PYQ — JEE Main 2023 (Apr 12 Shift 1)

Most Repeated PYQ — π f(sin x) Property