Wallis' Formula: A Complete Guide for JEE Aspirants

1. Introduction

Wallis' Formula, named after the English mathematician John Wallis (1616–1703), provides an efficient method for evaluating definite integrals of the form: $\int_0^{\pi/2} \sin^n x \, dx \quad \text{or} \quad \int_0^{\pi/2} \cos^n x \, dx,$ where $n$ is a non‑negative integer.
For JEE aspirants, mastering this formula saves valuable time in competitive exams by avoiding lengthy integration by parts or reduction‑formula steps. It is frequently applied directly or after a suitable substitution in JEE Main and Advanced problems.

---

2. Basic Symmetry

For any integer $n \ge 0$, $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx.$ This follows from the substitution $x \mapsto \frac{\pi}{2} - x$.
Denote the common value by $I_n$.

---

3. Wallis' Formula for $\int_0^{\pi/2} \sin^n x \, dx$

Case A: $n$ is odd ($n \ge 1$)

$I_n = \frac{(n-1)(n-3)(n-5) \cdots 4 \cdot 2}{\,n(n-2)(n-4) \cdots 5 \cdot 3\,}.$ Example: For $n = 7$,
$I_7 = \frac{6 \cdot 4 \cdot 2}{7 \cdot 5 \cdot 3} = \frac{48}{105} = \frac{16}{35}.$

Case B: $n$ is even ($n \ge 2$)

$I_n = \frac{(n-1)(n-3)(n-5) \cdots 3 \cdot 1}{\,n(n-2)(n-4) \cdots 4 \cdot 2\,} \cdot \frac{\pi}{2}.$ Example: For $n = 6$,
$I_6 = \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5\pi}{32}.$

Key points to remember:

• For odd $n$, the product in the numerator ends at 2, and the denominator ends at 3.
• For even $n$, the product in the numerator ends at 1, and the denominator ends at 2.
• The factor $\frac{\pi}{2}$ appears only when $n$ is even.

---

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

Let $J(m,n) = \int_0^{\pi/2} \sin^m x \cos^n x \, dx, \quad m, n \ge 0.$ Then $J(m,n) = \frac{\big[(m-1)(m-3)\cdots\big] \, \big[(n-1)(n-3)\cdots\big]} {(m+n)(m+n-2)(m+n-4)\cdots} \ \times\ \alpha,$ where:

• The products in the numerator run until a term $\le 1$ is reached.
• The product in the denominator decreases by 2 each step until a term $\le 2$ is reached.
• Factor $\alpha$ is:
  • $\alpha = \dfrac{\pi}{2}$ if both $m$ and $n$ are even;
  • $\alpha = 1$ otherwise.

Example 1: $m=4$ (even), $n=2$ (even)
$J(4,2) = \frac{(3\cdot1)(1)}{6\cdot4\cdot2} \cdot \frac{\pi}{2} = \frac{3}{48} \cdot \frac{\pi}{2} = \frac{\pi}{32}.$

Example 2: $m=3$ (odd), $n=4$ (even)
$J(3,4) = \frac{(2)(3\cdot1)}{7\cdot5\cdot3} \cdot 1 = \frac{6}{105} = \frac{2}{35}.$

---

5. Derivation via Reduction Formula

The formulas follow from the reduction relation $I_n = \frac{n-1}{n} I_{n-2}, \quad n \ge 2,$ with initial values $I_0 = \frac{\pi}{2}$ and $I_1 = 1$.
Repeated application gives the product formulas above.

For $J(m,n)$, the reduction formula is $J(m,n) = \frac{m-1}{m+n}\, J(m-2,n) = \frac{n-1}{m+n}\, J(m,n-2),$ used until the exponents become 0 or 1.

---

6. Quick Reference Table (Memorize for Speed)

---

7. Worked Examples (Step‑by‑Step)

Example 1

Evaluate $\displaystyle \int_0^{\pi/2} \sin^5 x \, dx$.

Solution:
Here $n = 5$ (odd).
$I_5 = \frac{4 \cdot 2}{5 \cdot 3} = \frac{8}{15}.$

Example 2

Evaluate $\displaystyle \int_0^{\pi/2} \cos^4 x \, dx$.

Solution:
Here $n = 4$ (even).
$I_4 = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}.$

Example 3

Evaluate $\displaystyle \int_0^{\pi/2} \sin^3 x \cos^5 x \, dx$.

Solution:
Here $m = 3$ (odd), $n = 5$ (odd) → $\alpha = 1$.
$J(3,5) = \frac{(2)(4\cdot2)}{8\cdot6\cdot4\cdot2} \cdot 1 = \frac{16}{384} = \frac{1}{24}.$

---

8. Previous Year JEE Problems (Solved)

Problem 1 (JEE Main Pattern)

Find $\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} \, dx$.

Solution:
Let $I = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} \, dx$.
Use the property $\int_{-a}^{a} f(x) dx = \int_0^a [f(x) + f(-x)] dx$.
Here $f(x) = \frac{\sin^2 x}{1+2^x}$, so $f(-x) = \frac{\sin^2 x}{1+2^{-x}} = \frac{2^x \sin^2 x}{1+2^x}.$ Adding, $f(x) + f(-x) = \sin^2 x.$ Hence $I = \int_0^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4} \quad (\text{from table}).$

Answer: $\dfrac{\pi}{4}$.

---

Problem 2 (JEE Advanced Pattern)

If $I_n = \int_0^{\pi/2} \sin^n x \, dx$, find $\dfrac{I_4 + I_6}{I_5 + I_7}$.

Solution:
From the table: $I_4 = \frac{3\pi}{16}, \quad I_5 = \frac{8}{15}, \quad I_6 = \frac{5\pi}{32}, \quad I_7 = \frac{16}{35}.$ Thus $I_4 + I_6 = \frac{3\pi}{16} + \frac{5\pi}{32} = \frac{6\pi + 5\pi}{32} = \frac{11\pi}{32},$ $I_5 + I_7 = \frac{8}{15} + \frac{16}{35} = \frac{56 + 48}{105} = \frac{104}{105}.$ Therefore $\frac{I_4 + I_6}{I_5 + I_7} = \frac{11\pi/32}{104/105} = \frac{11\pi \cdot 105}{32 \cdot 104} = \frac{1155\pi}{3328}.$

Answer: $\dfrac{1155\pi}{3328}$.

---

Problem 3 (IIT‑JEE Style)

Evaluate $\displaystyle \int_0^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$.

Hints:

1. Use $\sin^6 x + \cos^6 x = 1 - \frac{3}{4}\sin^2 2x$.
2. Substitute $t = 2x$ and simplify to an integral involving $\csc^2$ or use symmetry.
3. The final answer is $\frac{\pi}{2}$.
   (Try solving it completely as an exercise.)

---

9. Connection to Gamma Function

For advanced readers, Wallis' integrals can be expressed using the Gamma function: $\int_0^{\pi/2} \sin^n x \, dx = \frac{\sqrt{\pi}\; \Gamma\!\left(\frac{n+1}{2}\right)} {2\; \Gamma\!\left(\frac{n+2}{2}\right)}.$ This extends the formula to non‑integer $n$.

---

10. Wallis' Infinite Product for $\pi$

A famous corollary is Wallis' product: $\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \frac{2\cdot2}{1\cdot3} \cdot \frac{4\cdot4}{3\cdot5} \cdot \frac{6\cdot6}{5\cdot7} \cdots.$ It is historically important as an early infinite representation of $\pi$.

---

11. Essential Tips for JEE

1. Memorize the table up to $n = 8$.
2. Identify even/odd nature of the exponent quickly – the $\frac{\pi}{2}$ factor appears only for even $n$ in the basic formula, and only when both exponents are even in the general $J(m,n)$ formula.
3. Use symmetry properties first:
   • $\int_0^a f(x) dx = \int_0^a f(a-x) dx$
   • For even functions: $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$
   • For integrals with $\frac{1}{1+e^x}$ or similar, combine $f(x)$ and $f(-x)$.
4. Reduce to standard form via substitutions (e.g., $x = \frac{\pi}{2} - t$, or using double‑angle formulas).
5. Practice previous year problems to recognize when Wallis' formula is applicable, often after a simple manipulation.

---

12. Summary of Formulas

---

Master Wallis' formula through consistent practice. It is a powerful tool that, when combined with symmetry and substitution techniques, can solve many JEE definite‑integral problems in seconds.