Reduction Formulas — A Comprehensive Guide for JEE Main

What Are Reduction Formulas?

Reduction formulas are recursive relations that express an integral involving a power $n$ in terms of the same integral with a lower power $n-2$ or $n-1$. They systematically reduce complex integrals to simpler base cases, saving time and minimizing errors.

Core Reduction Formulas

1. For ∫ sinⁿx dx

$I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = -\cos x + C$

2. For ∫ cosⁿx dx

$I_n = \frac{\cos^{n-1}x \sin x}{n} + \frac{n-1}{n} I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = \sin x + C$

3. For ∫ tanⁿx dx

$I_n = \frac{\tan^{n-1}x}{n-1} - I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = \ln|\sec x| + C$

4. For ∫ cotⁿx dx

$I_n = -\frac{\cot^{n-1}x}{n-1} - I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = \ln|\sin x| + C$

5. For ∫ secⁿx dx

$I_n = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1} I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = \ln|\sec x + \tan x| + C$, $I_2 = \tan x + C$

6. For ∫ cosecⁿx dx

$I_n = -\frac{\csc^{n-2}x \cot x}{n-1} + \frac{n-2}{n-1} I_{n-2}$ Base cases:
$I_0 = x + C$, $I_1 = \ln|\csc x - \cot x| + C$, $I_2 = -\cot x + C$

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Solved Examples

Example 1: ∫ sin⁴x dx

Using the reduction formula for sinⁿx with $n = 4$: $I_4 = -\frac{\sin^3 x \cos x}{4} + \frac{3}{4} I_2$ For $I_2$: $I_2 = -\frac{\sin x \cos x}{2} + \frac{1}{2} I_0 = -\frac{\sin x \cos x}{2} + \frac{x}{2}$ Substituting back: $I_4 = -\frac{\sin^3 x \cos x}{4} + \frac{3}{4}\left(-\frac{\sin x \cos x}{2} + \frac{x}{2}\right)$ $\boxed{I_4 = -\frac{\sin^3 x \cos x}{4} - \frac{3\sin x \cos x}{8} + \frac{3x}{8} + C}$ Using $\sin 2x = 2\sin x \cos x$ and $\sin 4x = 8\sin^3 x \cos x - 4\sin x \cos x$, this simplifies to: $I_4 = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$

Example 2: ∫ sec⁵x dx

Using the reduction formula for secⁿx with $n = 5$: $I_5 = \frac{\sec^3 x \tan x}{4} + \frac{3}{4} I_3$ For $I_3$: $I_3 = \frac{\sec x \tan x}{2} + \frac{1}{2} I_1 = \frac{\sec x \tan x}{2} + \frac{1}{2}\ln|\sec x + \tan x|$ Thus: $\boxed{I_5 = \frac{\sec^3 x \tan x}{4} + \frac{3\sec x \tan x}{8} + \frac{3}{8}\ln|\sec x + \tan x| + C}$

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Wallis' Formulas for Definite Integrals

Standard Form

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

Extended Form (for ∫₀^(π/2) sinᵐx cosⁿx dx)

Let both $m$ and $n$ be non-negative integers. $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}$ For integer values: $= \frac{(m-1)!! (n-1)!!}{(m+n)!!} \times k$ where $k = \pi/2$ if both $m$ and $n$ are even, otherwise $k = 1$.

Notation: $n!!$ denotes the double factorial:

• For odd $n$: $n!! = n(n-2)(n-4)\cdots 1$
• For even $n$: $n!! = n(n-2)(n-4)\cdots 2$

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Wallis' Formula Examples

Example 3: ∫₀^(π/2) sin⁶x dx

$n = 6$ (even): $\int_0^{\pi/2} \sin^6 x \, dx = \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5\pi}{32}$

Example 4: ∫₀^(π/2) cos⁵x dx

$n = 5$ (odd): $\int_0^{\pi/2} \cos^5 x \, dx = \frac{4 \cdot 2}{5 \cdot 3} = \frac{8}{15}$

Example 5: ∫₀^(π/2) sin⁴x cos²x dx

$m = 4$ (even), $n = 2$ (even) → both even, so $k = \pi/2$

Numerator: $(3)(1) \times (1) = 3$

Denominator: $(6)(4)(2) = 48$

$\int_0^{\pi/2} \sin^4 x \cos^2 x \, dx = \frac{3}{48} \cdot \frac{\pi}{2} = \frac{\pi}{32}$

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Important Observations and Special Cases

Symmetry Properties

1. $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx$
2. $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \int_0^{\pi/2} \sin^n x \cos^m x \, dx$
3. $\int_0^{\pi} \sin^n x \, dx = 2\int_0^{\pi/2} \sin^n x \, dx$ for even $n$
4. $\int_0^{\pi} \cos^n x \, dx = \begin{cases} 0 & \text{if } n \text{ is odd} \\ 2\int_0^{\pi/2} \cos^n x \, dx & \text{if } n \text{ is even} \end{cases}$

Product-to-Sum Approach

For integrals like $\int \sin^m x \cos^n x \, dx$ with both $m$ and $n$ odd, it's often simpler to use: $\sin^m x \cos^n x = \frac{1}{2^{m+n-2}} \sum \text{terms of the form } \sin(kx) \text{ or } \cos(kx)$

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JEE Previous Year Questions

PYQ 1: JEE Main

If $I_n = \int_0^{\pi/4} \tan^n x \, dx$, then $I_n + I_{n+2} = \frac{1}{n+1}$.
Find $I_4 + I_6$.

Solution: From the given relation with $n = 4$: $I_4 + I_6 = \frac{1}{5}$

PYQ 2: JEE Advanced

If $I_n = \int_0^1 (1-x^5)^n dx$, prove that $I_n = \frac{5n}{5n+1} I_{n-1}$.
Find $\frac{I_6}{I_5}$.

Solution: Using integration by parts: $I_n = \int_0^1 (1-x^5)^n dx$ Let $u = (1-x^5)^n$, $dv = dx$. Then: $I_n = \left[x(1-x^5)^n\right]_0^1 + 5n \int_0^1 x^5 (1-x^5)^{n-1} dx$ Since $x^5 = 1 - (1-x^5)$: $I_n = 5n \int_0^1 [1 - (1-x^5)](1-x^5)^{n-1} dx = 5n(I_{n-1} - I_n)$ Solving: $I_n = \frac{5n}{5n+1} I_{n-1}$

For $n = 6$: $\frac{I_6}{I_5} = \frac{5 \times 6}{5 \times 6 + 1} = \frac{30}{31}$

PYQ 3: JEE Main 2021

Evaluate $\int_0^{\pi} x \sin^4 x \cos^6 x \, dx$.

Solution: Using property $\int_0^\pi x f(\sin x, \cos x) dx = \frac{\pi}{2} \int_0^\pi f(\sin x, \cos x) dx$ when $f$ is even about $\pi/2$: $I = \frac{\pi}{2} \int_0^\pi \sin^4 x \cos^6 x \, dx = \pi \int_0^{\pi/2} \sin^4 x \cos^6 x \, dx$ Using Wallis' extended formula with $m = 4$, $n = 6$ (both even): $\int_0^{\pi/2} \sin^4 x \cos^6 x \, dx = \frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{45\pi}{7680} = \frac{3\pi}{512}$ Thus: $I = \pi \cdot \frac{3\pi}{512} = \frac{3\pi^2}{512}$

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Common Pitfalls and Tips

1. Base Cases Matter: Always verify $I_0$ and $I_1$ for each reduction formula.
2. Sign Convention: Note the negative signs in sin, tan, cot, and cosec formulas.
3. Wallis' Formula Conditions: Remember the $\pi/2$ factor applies only when both exponents in $\int \sin^m x \cos^n x \, dx$ are even.
4. Symmetry Utilization: For definite integrals over symmetric intervals, check if the integrand has symmetry properties to simplify.
5. Alternative Methods: For small powers, direct expansion might be quicker than applying reduction formulas repeatedly.

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Practice Problems

1. Find $\int \cos^6 x \, dx$ using reduction formula.
2. Evaluate $\int_0^{\pi/2} \sin^7 x \, dx$ using Wallis' formula.
3. Find $\int \cot^4 x \, dx$.
4. Evaluate $\int_0^{\pi/2} \sin^3 x \cos^5 x \, dx$.
5. Derive reduction formula for $\int \sec^n x \, dx$.
6. If $I_{m,n} = \int_0^{\pi/2} \sin^m x \cos^n x \, dx$, prove that $I_{m,n} = \frac{m-1}{m+n} I_{m-2,n}$.
7. Evaluate $\int_0^{\pi} \sin^6 x \, dx$.

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Final Remarks

Reduction formulas and Wallis' formulas are powerful tools for JEE preparation. Remember:

• Use reduction formulas for indefinite integrals or when boundary conditions are not 0 to π/2.
• Use Wallis' formulas for definite integrals over [0, π/2] with trigonometric integrands.
• Always check if symmetry properties can simplify your calculation first.

Mastering these techniques will significantly enhance your efficiency in solving integration problems in JEE Main and Advanced.