King Property of Definite Integrals
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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.