# King Property of Definite Integrals

> King property for JEE Mathematics definite integrals: f(a-x) reflection, adding I to itself, x f(sin x) results and the log tan x integral equal to zero.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/definite-integral-king-property
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "King Property of Definite Integrals", PrepElephant, https://prepelephant.com/topics/jee/mathematics/definite-integral-king-property

## Direct answer

One reflection identity powers a whole family of JEE definite integrals: ∫[0 to a] f(x) dx = ∫[0 to a] f(a - x) dx, the king property, and its general form ∫[a to b] f(x) dx = ∫[a to b] f(a + b - x) dx. It works because the substitution x → a - x merely reverses the direction of traversal on a symmetric interval. Its force shows in the add-both-versions manoeuvre: write I in both forms, add, and the integrand often collapses — ∫[0 to π/2] dx/(1 + √tan x) = π/4 for any power, since adding f(x) + f(π/2 - x) gives 1. Corollaries worth memorising: ∫[0 to π] x f(sin x) dx = (π/2) ∫[0 to π] f(sin x) dx, and ∫[0 to π/2] log(tan x) dx = 0 while ∫[0 to π/2] log(sin x) dx = -(π/2) ln 2.

## What you must remember

- **The property:** ∫[0 to a] f(x) dx = ∫[0 to a] f(a - x) dx; the general interval version replaces a - x by a + b - x; the proof is the single substitution x = a - t.
- **Add-both-versions:** define I, rewrite via the king property, add the two equations — 2I then carries a simpler integrand; solve for I.
- **The tan-collapse:** ∫[0 to π/2] dx/(1 + tan^k x) = π/4 for every real k, because f(x) + f(π/2 - x) = 1 identically.
- **x times f(sin x):** ∫[0 to π] x f(sin x) dx = (π/2) ∫[0 to π] f(sin x) dx — the x disappears at the price of a factor.
- **Log integrals:** ∫[0 to π/2] log(tan x) dx = 0 (odd symmetry under the reflection) and ∫[0 to π/2] log(sin x) dx = ∫[0 to π/2] log(cos x) dx = -(π/2) ln 2, the Catalan-adjacent classic.
- **Even-odd splitting:** ∫[-a to a] f(x) dx = 0 for odd f and 2∫[0 to a] f(x) dx for even f — the partner symmetry in every toolbox.
- **Wallis reflection:** ∫[0 to π/2] sin^n x dx = ∫[0 to π/2] cos^n x dx — same property with f(x) = sin^n x and a = π/2.

## Adding I to itself, once cleanly

Evaluate I = ∫[0 to π/2] dx/(1 + √tan x). Direct integration is hopeless; the property is the question. Write the twin version by sending x → π/2 - x: √tan x becomes √cot x = 1/√tan x, so I = ∫[0 to π/2] dx/(1 + 1/√tan x) = ∫[0 to π/2] √tan x/(1 + √tan x) dx. Now add the two expressions for I: the integrands sum to [1 + √tan x]/(1 + √tan x) = 1, so 2I = ∫[0 to π/2] dx = π/2, giving I = π/4. Two habits make this reproducible under exam pressure: name the integral I before doing anything, and never actually compute the antiderivative — the whole point is that 2I carries the trivial integrand. The same skeleton solves ∫[0 to π/2] x/(sin x + cos x) dx (the x resolves through the x f(sin x) corollary after writing sin x + cos x = √2 sin(x + π/4)) and the log(tan x) integral, where the twin is the exact negative of the original, forcing 2I = 0.

## How the exam frames it

JEE Main loves this as a two-minute numerical: the tan-power integral equalling π/4, log sin x equalling -(π/2) ln 2, or an f(x) + f(a - x) sum given as data — the option π/4 or π/2 ln 2 recurring so often it is practically a house number. Advanced builds multi-layer versions: the property inside a function argument (f of an integral of f), chained with even-odd symmetry over [-π, π], or combined with periodicity where the interval splits into identical periods. The genuine slips: applying x → a - x to the integrand but forgetting the limits mirror too (they stay 0 and a — the reversal is automatic, and double-reversing by also flipping them is a classic error); assuming the property needs f continuous (a jump discontinuity inside still works if the pieces exist); and misremembering the log result's constant as π ln 2 or (π/2) ln 2 with the wrong sign. The king property is definite-integration syllabus gold for both papers.

## Frequently asked questions

### What is the king property of definite integrals?

∫[0 to a] f(x) dx = ∫[0 to a] f(a - x) dx, proved by the substitution x = a - t; on [a, b] it reads ∫ f(x) dx = ∫ f(a + b - x) dx.

### Why does ∫[0 to π/2] dx/(1 + tan^k x) equal π/4 for any k?

Adding the integral to its king-property twin makes the integrands sum to 1, so 2I = π/2 — the k never survives the addition.

### How is ∫[0 to π] x f(sin x) dx simplified?

To (π/2) ∫[0 to π] f(sin x) dx; the reflection pairs x with π - x, each carrying an average of π/2.

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

Zero — the reflected integrand log(cot x) = -log(tan x) makes the integral its own negative.

### Does the property hold if f has a discontinuity inside (0, a)?

Yes, as long as each piece is integrable — the substitution reverses the interval piecewise, and the values recombine unchanged.