Can I use Wallis's formula outside 0 to pi/2?

Not directly. For 0 to pi, use symmetry: the integral of sin to the n from 0 to pi equals twice the integral from 0 to pi/2. For cos to the n on 0 to pi, the answer is 0 if n is odd and twice the half-interval integral if n is even.

Why does the modulus inequality matter?

It is the integral analogue of the triangle inequality and gives an upper bound. Combined with the bounds property, if the absolute value of f is at most M, then the absolute value of the integral is at most M times the interval length.

Some Properties of Definite Integrals

Q: What is King's rule and when should I use it?

King's rule states that the integral of f(x) from a to b equals the integral of f(a + b - x) from a to b. Use it whenever the integrand contains trigonometric functions on intervals like 0 to pi/2 or 0 to pi, and adding the original and transformed integrals simplifies the sum.

Q: How do I remember Wallis's formula?

Numerator: start at (m-1) and step down by 2 until reaching 1 or 2; then start at (n-1) and step down. Denominator: start at (m+n) and step down by 2. Multiply by pi/2 only if both m and n are even.

Q: When does the even/odd property fail?

It requires the interval to be symmetric about zero (of the form [-a, a]). On any other interval it cannot be used. The function must also be genuinely even or odd.

Q: What if the integrand is neither even nor odd but the interval is symmetric?

Split the integrand into its even and odd parts. The odd part integrates to zero over a symmetric interval, leaving only the even part to compute.

Q: How is the estimation property useful in JEE problems?

Some JEE Advanced problems ask you to prove an integral lies between two given bounds without evaluating it. Sandwich the integrand between simpler functions using comparison and bounds properties, then integrate.

Q: What is the reduction formula and why do I need it?

A reduction formula expresses an integral in terms of a simpler one with a lower index. For sin to the n on 0 to pi/2, the recurrence I_n = (n-1)/n times I_{n-2} lets you compute high powers systematically. It underlies Wallis's formula.

Q: What is the periodic property and how do I detect the period?

If f(x + T) = f(x), the integral over any n periods is n times the integral over one period. For the absolute value of sine, the period is pi, not 2pi. Always find the smallest positive period before applying.

MathsIntegralsFor JEE aspirants

The properties of definite integrals are the working tools of the JEE definite-integration paper. They let you shortcut nasty evaluations that would take pages of direct substitution: King's rule $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , the even/odd property, the periodic property, and estimation inequalities. Mastering these ten properties, plus the reduction formulae and Wallis's formula, covers the majority of definite-integral problems asked in JEE Main and Advanced.

Key Formulas - Quick Reference

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ (dummy variable)
$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ (limit swap)
$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ (splitting)
King's rule: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$\int_{0}^{a} f (x) d x = \int_{0}^{a /2} [f (x) + f (a - x)] d x$
$\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f$ even; $= 0$ if $f$ odd
Periodic: $\int_{a}^{a + n T} f (x) d x = n \int_{0}^{T} f (x) d x$ if $f$ has period $T$
Bounds: if $m \leq f (x) \leq M$ on $[a, b]$ , then $m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$
$\int_{a}^{b} f (x) d x \leq \int_{a}^{b} ∣ f (x) ∣ d x$
Wallis: $\int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx = \begin{cases} \dfrac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2} \cdot \dfrac{\pi}{2}, & n \text{ even} \\[8pt] \dfrac{(n-1)(n-3)\cdots 2}{n(n-2)\cdots 3}, & n \text{ odd} \end{cases}$

1. The Ten Standard Properties

1Dummy variable

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t

The definite integral is independent of the variable of integration.

2Reversal of limits

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x

Swapping the upper and lower limits changes the sign of the integral.

3Splitting the interval

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x

Valid for any

c

(inside or outside

[a, b]

). Typically split first at points where

f (x)

is discontinuous, and second where the definition of

f (x)

changes (modulus, greatest integer, piecewise).

Solved Example 1

Evaluate

\int_{0}^{5 π /12} [tan x] d x

, where

[\cdot]

is the greatest integer function.

Solution:

At $x = 0$ : $tan x = 0$ . At $x = 5 π /12$ : $tan (5 π /12) = 2 + 3 \approx 3.73$ .

The integers taken by $[tan x]$ on $[0, 5 π /12]$ are $0, 1, 2, 3$ . The transitions occur where $tan x = 1, 2, 3$ , i.e., $x = tan^{- 1} 1, tan^{- 1} 2, tan^{- 1} 3$ .

$I = \int_{0}^{t a n^{- 1} 1} 0 d x + \int_{t a n^{- 1} 1}^{t a n^{- 1} 2} 1 d x + \int_{t a n^{- 1} 2}^{t a n^{- 1} 3} 2 d x + \int_{t a n^{- 1} 3}^{5 π /12} 3 d x$

$= 0 + (tan^{- 1} 2 - tan^{- 1} 1) + 2 (tan^{- 1} 3 - tan^{- 1} 2) + 3 (\frac{5 π}{12} - tan^{- 1} 3)$

$= \frac{5 π}{4} - tan^{- 1} 1 - tan^{- 1} 2 - tan^{- 1} 3$

Using $tan^{- 1} 2 + tan^{- 1} 3 = π - tan^{- 1} 1 = π - π /4 = 3 π /4$ :

$I = \frac{5 π}{4} - \frac{π}{4} - \frac{3 π}{4} = \frac{π}{4}$

4King's rule

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

In particular,

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

. This is the single most useful property for JEE trigonometric integrals.

Solved Example 2

Prove that

\int_{0}^{π /2} \frac{g ( sin x )}{g ( sin x ) + g ( cos x )} d x = \frac{π}{4}

Solution:

Let $I = \int_{0}^{π /2} \frac{g ( sin x )}{g ( sin x ) + g ( cos x )} d x$ . Apply King's rule with $a = 0$ , $b = π /2$ , so $x \to π /2 - x$ :

$I = \int_{0}^{π /2} \frac{g ( sin ( π /2 - x ))}{g ( sin ( π /2 - x )) + g ( cos ( π /2 - x ))} d x = \int_{0}^{π /2} \frac{g ( cos x )}{g ( cos x ) + g ( sin x )} d x$

Adding the two expressions:

$2 I = \int_{0}^{π /2} \frac{g ( sin x ) + g ( cos x )}{g ( sin x ) + g ( cos x )} d x = \int_{0}^{π /2} 1 d x = \frac{π}{2}$

Hence $I = π /4$ .

Solved Example 3

f, g, h

are continuous on

[0, a]

satisfying

f (a - x) = f (x)

g (a - x) = - g (x)

, and

3 h (x) - 4 h (a - x) = 5

, prove that

\int_{0}^{a} f (x) g (x) h (x) d x = 0

Solution:

Let $I = \int_{0}^{a} f (x) g (x) h (x) d x$ . By King's rule:

$I = \int_{0}^{a} f (a - x) g (a - x) h (a - x) d x = - \int_{0}^{a} f (x) g (x) h (a - x) d x$

So $I = - \int_{0}^{a} f g h (a - x) d x$ . Multiply the first form by 3 and the second by 4:

$3 I = 3 \int_{0}^{a} f g h (x) d x, 4 I = - 4 \int_{0}^{a} f g h (a - x) d x$

Adding: $7 I = \int_{0}^{a} f g [3 h (x) - 4 h (a - x)] d x = \int_{0}^{a} f g \cdot 5 d x = 5 \int_{0}^{a} f g d x$ .

But $f (a - x) g (a - x) = - f (x) g (x)$ , so $\int_{0}^{a} f g d x = - \int_{0}^{a} f g d x$ , giving $\int_{0}^{a} f g d x = 0$ . Hence $7 I = 0$ , so $I = 0$ .

Solved Example 4

Evaluate

I = \int_{0}^{π} x sin 2 x sin (\frac{π}{2} cos x) d x

Solution:

By King's rule with $x \to π - x$ :

$I = \int_{0}^{π} (π - x) sin (2 π - 2 x) sin (\frac{π}{2} cos (π - x)) d x$

$= \int_{0}^{π} (π - x) (- sin 2 x) sin (- \frac{π}{2} cos x) d x = \int_{0}^{π} (π - x) sin 2 x sin (\frac{π}{2} cos x) d x$

Adding to the original:

$2 I = π \int_{0}^{π} sin 2 x sin (\frac{π}{2} cos x) d x ⟹ I = π \int_{0}^{π /2} 2 sin x cos x sin (\frac{π}{2} cos x) d x$

Substitute $z = \frac{π}{2} cos x$ , $d z = - \frac{π}{2} sin x d x$ . When $x = 0$ , $z = π /2$ ; when $x = π /2$ , $z = 0$ :

$I = π \int_{π /2}^{0} 2 \cdot \frac{2 z}{π} \cdot (- \frac{2}{π}) sin z d z = \frac{8}{π} \int_{0}^{π /2} z sin z d z$

Integration by parts: $\int z sin z d z = - z cos z + sin z$ . Evaluated on $[0, π /2]$ : $[- z cos z + sin z]_{0}^{π /2} = 0 + 1 - 0 = 1$ .

Hence $I = \frac{8}{π}$ .

5Half-interval reduction

\int_{0}^{a} f (x) d x = \int_{0}^{a /2} [f (x) + f (a - x)] d x

Special cases:

If $f (x) = f (a - x)$ : $\int_{0}^{a} f (x) d x = 2 \int_{0}^{a /2} f (x) d x$
If $f (x) = - f (a - x)$ : $\int_{0}^{a} f (x) d x = 0$

6Even and odd functions

\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x

Special cases:

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

Figure 1: Even functions double the half-interval area; odd functions integrate to zero on symmetric intervals

[- a, a]

Solved Example 5

Evaluate

\int_{- 4}^{4} \frac{x ^{2}}{( x ^{2} + 16 ) ( 1 + e ^{x^{5}} )} d x

Solution:

Let $f (x) = \frac{x ^{2}}{x ^{2} + 16}$ , so the integrand is $\frac{f ( x )}{1 + e ^{x^{5}}}$ . Note $f$ is even. Denote the integral by $I$ :

$I = \int_{- 4}^{4} \frac{f ( x )}{1 + e ^{x^{5}}} d x$

By King's rule with $x \to - x$ on the symmetric interval:

$I = \int_{- 4}^{4} \frac{f ( - x )}{1 + e ^{(- x)^{5}}} d x = \int_{- 4}^{4} \frac{f ( x )}{1 + e ^{- x^{5}}} d x = \int_{- 4}^{4} \frac{f ( x ) e ^{x^{5}}}{1 + e ^{x^{5}}} d x$

Adding: $2 I = \int_{- 4}^{4} f (x) d x = 2 \int_{0}^{4} \frac{x ^{2}}{x ^{2} + 16} d x$ (since $f$ is even).

$I = \int_{0}^{4} \frac{x ^{2}}{x ^{2} + 16} d x = \int_{0}^{4} (1 - \frac{16}{x ^{2} + 16}) d x = [x - 4 tan^{- 1} \frac{x}{4}]_{0}^{4} = 4 - π$

Solved Example 6

Find the value of

\int_{0}^{π} \frac{d x}{1 + 5 ^{c o s x}} + \int_{- 2}^{2} lo g (\frac{5 - x}{5 + x}) d x

Solution:

Call these $I_{1}$ and $I_{2}$ .

For $I_{1}$ : Apply King's rule with $x \to π - x$ :

$I_{1} = \int_{0}^{π} \frac{d x}{1 + 5 ^{- c o s x}} = \int_{0}^{π} \frac{5 ^{c o s x} d x}{5 ^{c o s x} + 1}$

Adding the two forms: $2 I_{1} = \int_{0}^{π} 1 d x = π$ , so $I_{1} = π /2$ .

For $I_{2}$ : Let $g (x) = lo g (\frac{5 - x}{5 + x})$ . Then $g (- x) = lo g (\frac{5 + x}{5 - x}) = - g (x)$ , so $g$ is odd. Since the interval $[- 2, 2]$ is symmetric, $I_{2} = 0$ .

Hence total $= π /2 + 0 = \frac{π}{2}$ .

7Scale to

[0, 1]

\int_{a}^{b} f (x) d x = (b - a) \int_{0}^{1} f ((b - a) x + a) d x

Useful when you can spot a mirror-image integral over a different interval.

Solved Example 7

Evaluate

\int_{- 4}^{- 5} e^{(x + 5)^{2}} d x + 3 \int_{1/3}^{2/3} e^{9 (x - 2/3)^{2}} d x

Solution:

Let $I_{1} = \int_{- 4}^{- 5} e^{(x + 5)^{2}} d x$ . Note the limits are reversed. Apply Property 7 with $a = - 4$ , $b = - 5$ , so $b - a = - 1$ :

$I_{1} = - \int_{0}^{1} e^{(- x + 1)^{2}} d x = - \int_{0}^{1} e^{(x - 1)^{2}} d x$

Let $I_{2} = \int_{1/3}^{2/3} e^{9 (x - 2/3)^{2}} d x$ . With $a = 1/3$ , $b = 2/3$ , so $b - a = 1/3$ :

$I_{2} = \frac{1}{3} \int_{0}^{1} e^{9 (x /3 + 1/3 - 2/3)^{2}} d x = \frac{1}{3} \int_{0}^{1} e^{(x - 1)^{2}} d x = - \frac{I _{1}}{3}$

Total $= I_{1} + 3 I_{2} = I_{1} + 3 \cdot (- I_{1} /3) = I_{1} - I_{1} = 0$ .

8Periodic functions

f (x)

has period

T

and

n \in Z

, then:

\int_{a}^{a + n T} f (x) d x = n \int_{0}^{T} f (x) d x

Special cases:

$\int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x$
$\int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x$ (translation invariance)

Solved Example 8

Evaluate

\int_{0}^{10 π} ∣ sin x ∣ d x

Solution:

$∣ sin x ∣$ is periodic with period $π$ . Hence:

$\int_{0}^{10 π} ∣ sin x ∣ d x = 10 \int_{0}^{π} ∣ sin x ∣ d x = 10 \int_{0}^{π} sin x d x = 10 [- cos x]_{0}^{π} = 10 \cdot 2 = 20$

Solved Example 9

f (x)

satisfies

f (x + a) + f (x) = 0

for all

x \in R

and constant

a

, and

\int_{b}^{c + b} f (x) d x

is independent of

b

, find the least positive value of

c

Solution:

From $f (x + a) = - f (x)$ , replace $x$ by $x + a$ : $f (x + 2 a) = - f (x + a) = f (x)$ . So $f$ is periodic with period $2 a$ .

The condition " $\int_{b}^{c + b} f (x) d x$ independent of $b$ " holds precisely when $c$ is an integer multiple of the period. The least positive value is $c = 2 a$ .

Properties 9–12: Bounds and inequalities

9Comparison

ψ (x) \leq f (x) \leq ϕ (x)

for

a \leq x \leq b

, then

\int_{a}^{b} ψ (x) d x \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} ϕ (x) d x

10Bounds via extrema

m \leq f (x) \leq M

for

a \leq x \leq b

, then

m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)

11Absolute value inequality

\int_{a}^{b} f (x) d x \leq \int_{a}^{b} ∣ f (x) ∣ d x

The integral analogue of the triangle inequality.

12Sign preservation

f (x) \geq 0

[a, b]

, then

\int_{a}^{b} f (x) d x \geq 0

For monotonic

f

(a, b)

: if

f

is decreasing,

f (b) (b - a) < \int_{a}^{b} f (x) d x < f (a) (b - a)

; if increasing, the reverse.

Solved Example 10

Estimate the value of

\int_{0}^{π /2} \frac{sin x}{x} d x

Solution:

Let $f (x) = \frac{sin x}{x}$ . Then $f^{'} (x) = \frac{x cos x - sin x}{x ^{2}} = \frac{cos x ( x - tan x )}{x ^{2}} < 0$ on $(0, π /2)$ , since $tan x > x$ there.

So $f$ is monotonically decreasing. As $x \to 0^{+}$ , $f (x) \to 1$ . At $x = π /2$ : $f (π /2) = \frac{2}{π}$ .

By Property 10 (bounds): $m = 2/ π$ , $M = 1$ , $b - a = π /2$ :

$\frac{2}{π} \cdot \frac{π}{2} < \int_{0}^{π /2} \frac{sin x}{x} d x < 1 \cdot \frac{π}{2} ⟹ 1 < \int_{0}^{π /2} \frac{sin x}{x} d x < \frac{π}{2}$

Solved Example 11

Prove that

\frac{π}{6} < \int_{0}^{1} \frac{d x}{4 - x ^{2} - x ^{3}} < \frac{π}{4 2}

Solution:

For $x \in (0, 1)$ : $0 < x^{3} < x^{2}$ , so $x^{2} < x^{2} + x^{3} < 2 x^{2}$ , hence $- 2 x^{2} < - x^{2} - x^{3} < - x^{2}$ .

Therefore $4 - 2 x^{2} < 4 - x^{2} - x^{3} < 4 - x^{2}$ , giving $4 - 2 x^{2} < 4 - x^{2} - x^{3} < 4 - x^{2}$ .

Taking reciprocals reverses the inequalities:

$\frac{1}{4 - x ^{2}} < \frac{1}{4 - x ^{2} - x ^{3}} < \frac{1}{4 - 2 x ^{2}}$

Integrating on $[0, 1]$ :

$[sin^{- 1} \frac{x}{2}]_{0}^{1} < I < [\frac{1}{2} sin^{- 1} \frac{x}{2}]_{0}^{1} ⟹ \frac{π}{6} < I < \frac{π}{4 2}$

2. Reduction Formulae

Reduction formulae express $\int f_{n} (x) d x$ in terms of $\int f_{n - 2} (x) d x$ (or similar), allowing systematic evaluation of trigonometric powers.

Reduction formula for $sin^{n} x$

I_{n} = \int_{0}^{π /2} sin^{n} x d x

, then

I_{n} = \frac{n - 1}{n} I_{n - 2}

with

I_{0} = \frac{π}{2}

and

I_{1} = 1

Proof outline: Write $sin^{n} x = sin^{n - 1} x \cdot sin x$ and apply integration by parts with $u = sin^{n - 1} x$ , $d v = sin x d x$ :

$I_{n} = [- sin^{n - 1} x cos x]_{0}^{π /2} + (n - 1) \int_{0}^{π /2} sin^{n - 2} x cos^{2} x d x$

The boundary term vanishes. Using $cos^{2} x = 1 - sin^{2} x$ : $I_{n} = (n - 1) I_{n - 2} - (n - 1) I_{n}$ , so $n I_{n} = (n - 1) I_{n - 2}$ .

Reduction formula for $tan^{n} x$

I_{n} = \int_{0}^{π /4} tan^{n} x d x

, then

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

Proof: Write $tan^{n} x = tan^{n - 2} x (sec^{2} x - 1)$ :

$I_{n} = \int_{0}^{π /4} tan^{n - 2} x sec^{2} x d x - I_{n - 2} = [\frac{tan ^{n - 1} x}{n - 1}]_{0}^{π /4} - I_{n - 2} = \frac{1}{n - 1} - I_{n - 2}$

Reduction formula for $sin^{m} x cos^{n} x$

I_{m, n} = \int_{0}^{π /2} sin^{m} x cos^{n} x d x

, then

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

Symmetrically, one can reduce in

n

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

3. Wallis's Formula

For non-negative integers

m, n

I_{m, n} = \int_{0}^{π /2} sin^{m} x cos^{n} x d x = \frac{( m - 1 ) ( m - 3 ) \dots \cdot ( n - 1 ) ( n - 3 ) \dots}{( m + n ) ( m + n - 2 ) ( m + n - 4 ) \dots} \cdot k

where $k = \frac{π}{2}$ if both $m$ and $n$ are even; $k = 1$ otherwise.

Each product runs down to $2$ or $1$ ; stop at $1$ if the last factor is even (i.e. when $m - 1$ or $n - 1$ reaches $0$ ).

Special case: $\int_{0}^{π /2} sin^{n} x d x$

\int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx = \begin{cases} \dfrac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2} \cdot \dfrac{\pi}{2}, & n \text{ even} \\[8pt] \dfrac{(n-1)(n-3)\cdots 2}{n(n-2)\cdots 3}, & n \text{ odd} \end{cases}

Solved Example 12

Evaluate

\int_{- π /2}^{π /2} sin^{2} x cos^{2} x (sin x + cos x) d x

Solution:

Expand: $\int_{- π /2}^{π /2} sin^{3} x cos^{2} x d x + \int_{- π /2}^{π /2} sin^{2} x cos^{3} x d x$ .

The first integrand is odd ( $sin^{3} x cos^{2} x$ : odd times even), so its integral over the symmetric interval is $0$ . The second is even, so it doubles the half-interval integral:

$= 0 + 2 \int_{0}^{π /2} sin^{2} x cos^{3} x d x$

By Wallis: $m = 2, n = 3$ (not both even), so $k = 1$ :

$\int_{0}^{π /2} sin^{2} x cos^{3} x d x = \frac{( 2 - 1 ) ( 3 - 1 )}{( 5 ) ( 3 ) ( 1 )} = \frac{1 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{2}{15}$

Total $= 2 \cdot \frac{2}{15} = \frac{4}{15}$ .

Solved Example 13

Evaluate

\int_{0}^{π} x sin^{5} x cos^{6} x d x

Solution:

Let $I = \int_{0}^{π} x sin^{5} x cos^{6} x d x$ . Apply King's rule with $x \to π - x$ :

$I = \int_{0}^{π} (π - x) sin^{5} x cos^{6} x d x \cdot (- 1)^{6} \cdot (- 1)^{5} \cdot (- 1)^{5}$

More cleanly: $sin (π - x) = sin x$ , $cos (π - x) = - cos x$ , so $sin^{5} (π - x) cos^{6} (π - x) = sin^{5} x \cdot cos^{6} x$ .

$I = \int_{0}^{π} (π - x) sin^{5} x cos^{6} x d x = π \int_{0}^{π} sin^{5} x cos^{6} x d x - I$

So $2 I = π \int_{0}^{π} sin^{5} x cos^{6} x d x$ .

The integrand on $[0, π]$ : substituting $x \to π - x$ multiplies $cos^{6}$ by $(- 1)^{6} = 1$ and $sin^{5}$ by $1$ , so integrand is symmetric about $x = π /2$ . Hence $\int_{0}^{π} = 2 \int_{0}^{π /2}$ :

$2 I = 2 π \int_{0}^{π /2} sin^{5} x cos^{6} x d x$

By Wallis ( $m = 5, n = 6$ , not both even): $\int_{0}^{π /2} sin^{5} x cos^{6} x d x = \frac{4 \cdot 2 \cdot 5 \cdot 3 \cdot 1}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{8}{693}$ .

Therefore $I = π \cdot \frac{8}{693} = \frac{8 π}{693}$ .

Common Mistakes to Avoid

Watch out

Applying the even/odd property when the interval is not symmetric about zero. It only works on $[- a, a]$ .
Forgetting to check the sign of $cos^{n} (π - x) = (- 1)^{n} cos^{n} x$ when using King's rule with a $π$ -interval. The parity of $n$ matters.
Confusing periodic reduction ( $\int_{0}^{n T} = n \int_{0}^{T}$ ) with the general splitting. Property 8 requires periodicity.
Miscounting products in Wallis's formula. The numerator uses $(m - 1) (m - 3) \dots$ and $(n - 1) (n - 3) \dots$ ; the denominator uses $(m + n) (m + n - 2) \dots$ .
Forgetting the $π /2$ factor in Wallis when both $m$ and $n$ are even. If either is odd, no $π /2$ .
Applying inequality manipulations (like reciprocals) without noting the direction reversal for positive quantities.
Using King's rule with the wrong end-point sum. It's $f (a + b - x)$ , not $f (b - x)$ unless $a = 0$ .

Frequently Asked Questions

Q1. What is King's rule and when should I use it?

King's rule (Property 4) states $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ . Use it whenever the integrand contains $sin x$ , $cos x$ , $tan x$ , or a rational expression of them on $[0, π /2]$ or $[0, π]$ , and adding the original and King-transformed integrals simplifies the sum.

Q2. How do I remember Wallis's formula?

Numerator: start at $(m - 1)$ and step down by 2 until reaching 1 or 2; then start at $(n - 1)$ and step down. Denominator: start at $(m + n)$ and step down by 2 until reaching 1 or 2. Multiply by $π /2$ only if both $m$ and $n$ are even.

Q3. When does the even/odd property fail?

It requires the interval to be symmetric about zero (of the form $[- a, a]$ ). On any other interval, you cannot use it. Also, the function itself must be even or odd; many trigonometric expressions are neither.

Q4. What if the integrand is neither even nor odd but the interval is $[- a, a]$ ?

Split the integrand into its even and odd parts: $f (x) = \frac{f ( x ) + f ( - x )}{2} + \frac{f ( x ) - f ( - x )}{2}$ . The odd part integrates to zero over $[- a, a]$ , leaving only the even part to compute.

Q5. How is the estimation property useful in JEE problems?

Some JEE Advanced problems ask you to prove an integral lies between two given values without evaluating it. Property 9 (comparison) and Property 10 (bounds via extrema) are the standard tools. Sandwich the integrand between simpler functions, then integrate the bounds.

Q6. What is the reduction formula and why do I need it?

A reduction formula expresses an integral in terms of a simpler one with a lower index. For $\int_{0}^{π /2} sin^{n} x d x$ , the recurrence $I_{n} = \frac{n - 1}{n} I_{n - 2}$ lets you compute high powers systematically. Reduction formulae are the mechanism behind Wallis's formula.

Q7. Can I use Wallis's formula outside $[0, π /2]$ ?

Not directly. Wallis's formula is stated for $[0, π /2]$ . For $[0, π]$ , use symmetry: $\int_{0}^{π} sin^{n} x d x = 2 \int_{0}^{π /2} sin^{n} x d x$ . For $cos^{n} x$ on $[0, π]$ , the answer is $0$ if $n$ is odd (by King's rule) and $2 \int_{0}^{π /2} cos^{n} x d x$ if $n$ is even.

Q8. What is the periodic property and how do I detect the period?

If $f (x + T) = f (x)$ , then $\int_{a}^{a + n T} f (x) d x = n \int_{0}^{T} f (x) d x$ . For $∣ sin x ∣$ the period is $π$ , not $2 π$ . For $sin^{2} x$ , the period is also $π$ . Always find the smallest positive period before applying.

Q9. Why does the modulus inequality $∣ \int f ∣ \leq \int ∣ f ∣$ matter?

It's the integral analogue of the triangle inequality and gives an upper bound on the size of an integral. If $∣ f (x) ∣ \leq M$ , then $∣ \int_{a}^{b} f (x) d x ∣ \leq \int_{a}^{b} ∣ f (x) ∣ d x \leq M (b - a)$ - a two-step estimate combining Properties 10 and 11.

Previous year questions on Some Properties of Definite Integrals

23 questions from past papers, each with a step-by-step solution.

Show all 23 questions

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1. The Ten Standard Properties

Properties 9–12: Bounds and inequalities

2. Reduction Formulae

Reduction formula for sinnx

Reduction formula for tannx

Reduction formula for sinmxcosnx

3. Wallis's Formula

Special case: ∫0π/2​sinnxdx

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is King's rule and when should I use it?

Q2. How do I remember Wallis's formula?

Q3. When does the even/odd property fail?

Q4. What if the integrand is neither even nor odd but the interval is [−a,a]?

Q5. How is the estimation property useful in JEE problems?

Q6. What is the reduction formula and why do I need it?

Q7. Can I use Wallis's formula outside [0,π/2]?

Q8. What is the periodic property and how do I detect the period?

Q9. Why does the modulus inequality ∣∫f∣≤∫∣f∣ matter?

Previous year questions on Some Properties of Definite Integrals

Ready to master Integrals?

Reduction formula for $sin^{n} x$

Reduction formula for $tan^{n} x$

Reduction formula for $sin^{m} x cos^{n} x$

Special case: $\int_{0}^{π /2} sin^{n} x d x$

Q4. What if the integrand is neither even nor odd but the interval is $[- a, a]$ ?

Q7. Can I use Wallis's formula outside $[0, π /2]$ ?

Q9. Why does the modulus inequality $∣ \int f ∣ \leq \int ∣ f ∣$ matter?