# Definite Integral Evaluation Tricks

> Definite integral evaluation tricks for JEE Mathematics: king property, odd-even pairing, x f(sin x) rule and the ln(sin x) = −(π/2) ln 2 classic.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/definite-integral-evaluation-tricks
- 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: "Definite Integral Evaluation Tricks", PrepElephant, https://prepelephant.com/topics/jee/mathematics/definite-integral-evaluation-tricks

## Direct answer

Limits are information. The king property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx lets an integral meet its own reflection, and when the two versions simplify — f(x) + f(a − x) constant, or logarithms pairing into products — the integral solves itself: I = ∫₀^{π/2} ln(sin x) dx famously equals −(π/2) ln 2 by exactly this self-meeting. The supporting cast: odd functions integrate to zero over symmetric intervals, even ones double; ∫₀^{π/2} f(sin x) dx = ∫₀^{π/2} f(cos x) dx; and ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx, which deletes any x standing beside a sine. Every trick is a substitution chosen before hunting antiderivatives — the skill is choosing it.

## What you must remember

- **King property:** ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx; adding the two forms and simplifying f(x) + f(a − x) finishes most items in two lines.
- **Symmetry:** odd integrand on [−a, a] gives zero; even gives twice the half-interval integral.
- **Sine-cosine exchange:** on [0, π/2], sin may be replaced by cosine throughout.
- **The x-killer:** ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx, the trick behind every "x sin x/(1 + cos²x)" item.
- **The logarithmic classic:** ∫₀^{π/2} ln(sin x) dx = −(π/2) ln 2, and by exchange the cosine version matches it.
- **Rescaling discipline:** under t = 2x the limits halve in number and a factor one-half appears — forgotten factors are the commonest numerical error.
- **Periodicity shift:** for period T, ∫₀^{nT} f = n∫₀^{T} f, and ∫₀^{T} f(x) dx = ∫ₐ^{a+T} f(x) dx for any a.

## The logarithmic sine integral

Evaluate I = ∫₀^{π/2} ln(sin x) dx. By the king property with a = π/2, I = ∫₀^{π/2} ln(cos x) dx as well. Add the two equations: 2I = ∫₀^{π/2} ln(sin x cos x) dx = ∫₀^{π/2} ln(sin 2x/2) dx = ∫₀^{π/2} ln(sin 2x) dx − (π/2) ln 2. Substitute t = 2x in the remaining integral: it becomes (1/2)∫₀^π ln(sin t) dt, and since sin t is symmetric about π/2 on [0, π], this equals (1/2) × 2∫₀^{π/2} ln(sin t) dt = I. So 2I = I − (π/2) ln 2, giving I = −(π/2) ln 2 ≈ −1.0888 — a negative value, as the graph of ln(sin x) below the axis demands. The structure to internalise: the substitution did not evaluate the integral, it recognised the integral inside itself, and the equation solved for I algebraically. That self-referential loop is the signature of the hardest definite-integral items in the paper.

## Tricks versus traps

JEE Main runs the one-shot king property: ∫₀^π x sin x/(1 + cos²x) dx becomes (π/2)∫₀^π sin x/(1 + cos²x) dx = (π/2)[−arctan(cos x)]₀^π = (π/2)(arctan 1 + arctan 1) = π²/4 — a known numerical answer worth rehearsing end to end. The distractors are π²/2 (factor dropped) and π/4 (π forgotten). JEE Advanced chains two properties or poses I − J systems, where two integrals are defined and their sum and difference are both computed by substitution — solving simultaneous equations with integrals. The trap inventory: after t = 2x, forgetting the 1/2 factor; claiming ∫₀^{2π} f(sin x) dx = 2∫₀^π f(sin x) dx for any f (true only because sin is symmetric on each half — check, do not assume); and pairing ln terms before confirming both integrals converge. The professional order of operations: symmetry check, king property, x-killer, only then antiderivative — reversing that order is how candidates spend eight minutes on a two-line item.

## Frequently asked questions

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

∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx; adding the two forms often collapses the integrand to a constant or a product identity.

### What is the value of the integral of ln(sin x) from 0 to π/2?

−(π/2) ln 2, obtained by pairing with the cosine version and substituting t = 2x.

### How does the x f(sin x) rule work?

∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx, so the factor x disappears at the cost of multiplying by π/2.

### What happens to an odd function on a symmetric interval?

It integrates to exactly zero on [−a, a]; an even integrand gives double the integral over [0, a].

### What must be checked when substituting t = 2x inside a definite integral?

Both the limits (0 to π) and the factor 1/2 from dx = dt/2 — the two steps where most numerical answers die.