NDA Maths · Definite Integration

Properties — King's, Symmetry and Standard Results

The properties of definite integrals — above all King's property and odd/even symmetry — evaluate integrals you could never antidifferentiate, by adding an integral to its own mirror image.

All Definite Integration notes NDA Maths strategy

Why this matters

This is the heart of the chapter: 32 PYQs and most of its HARD questions. One technique dominates — King's property, the 'replace x by (a−x) and add' move that collapses a fearsome integrand into a constant. Add odd/even symmetry, a short list of standard results, and you can do almost every Properties question by inspection.

Concept 1 of 4

King's property — the reflection trick

Intuition

The single most powerful definite-integral move: an integral from 0 to a is unchanged if you replace x by (a−x). When the integrand has a symmetric structure, calling the integral I and adding it to its reflected form makes the messy parts cancel, leaving a trivial integral — you get 2I = something easy.

Definition

King's property and the 2I technique:

King's property: $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ (and $\int_b^c f(x)\,dx=\int_b^c f(b+c-x)\,dx$ ).
The 2I move: let $I=\int_0^a f(x)\,dx$ ; write a second copy with $x\to a-x$ ; ADD them so the denominators match. Often $2I=\int_0^a 1\,dx = a$ .
Classic family: $\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}$ .
Works on $\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx = \frac{\pi}{4}$ and the whole $\frac{a\cos x+b\sin x}{\sin x+\cos x}$ family.

King's property and its companions

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \qquad \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \qquad \int_0^{2a} f(x)\,dx = \int_0^a \big[f(x)+f(2a-x)\big]\,dx \qquad \int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}

Worked example

Evaluate

I=\displaystyle\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Definite IntegrationMODERATE

What is

\displaystyle\int_0^{\pi/2}\dfrac{a+\sin x}{2(a+\sin x+\cos x)}\,dx

equal to?

[Q78 · Apr · 2024]

Add the reflected form — don't just substitute and stop

King's property by itself only rewrites

I

; the magic is ADDING the original and the reflected integral so the numerator becomes the denominator. If you substitute and forget to add, you go in a circle.

King's reflection only works with the matching limits

The reflection is

x\to a+b-x

for limits

[a,b]

— NOT a fixed

x\to a-x

. On

\int_2^5 f\,dx

you must replace

x

7-x

, not

2-x

. Using the wrong reflection sends the limits outside the interval and the cancellation fails.

Drill 13 more on king's property — the reflection trick

Concept 2 of 4

Odd/even symmetry over a symmetric interval

Intuition

On an interval symmetric about zero, an odd function integrates to zero (the left half cancels the right) and an even function integrates to twice the half. A bonus property handles integrands divided by 1 plus an exponential.

Definition

Symmetry rules on $[-a,a]$ :

Odd $f(-x)=-f(x)$ : $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ .
Even $f(-x)=f(x)$ : $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$ .
**The $1+c^x$ property**: for even $f$ , $\displaystyle\int_{-a}^{a}\frac{f(x)}{1+c^{x}}\,dx = \frac12\int_{-a}^{a} f(x)\,dx = \int_0^a f(x)\,dx$ .
Also $\displaystyle\int_{-a}^{a} h(x)\,dx = \int_0^a [h(x)+h(-x)]\,dx$ .

Odd/even symmetry over a symmetric interval

\int_{-a}^{a} f(x)\,dx = 0 \;\;(f\text{ odd}) \qquad \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \;\;(f\text{ even}) \qquad \int_{-a}^{a} \frac{f(x)}{1+c^{x}}\,dx = \int_0^a f(x)\,dx \;\;(f\text{ even})

Worked example

Evaluate

\displaystyle\int_{-\pi/4}^{\pi/4} (\sin x - \tan x)\,dx

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 2Definite IntegrationMODERATE

What is

\int_{-\pi/2}^{\pi/2}(e^{\cos x}\sin x+e^{\sin x}\cos x)\,dx

equal to?

[Q96 · Sep · 2022]

Check parity of the WHOLE integrand

A sum can have one odd part and one non-odd part. Split it: the odd piece dies, but the rest must still be integrated. Don't declare the whole integral zero just because one term is odd.

Drill 6 more on odd/even symmetry over a symmetric interval

Concept 3 of 4

Standard results and trig reductions

Intuition

A handful of definite integrals recur so often they are worth knowing as results, and a few power-reduction identities turn an un-integrable trig power into something elementary. Recognising these saves the whole derivation.

Definition

Results and reductions to memorise:

$\displaystyle\int_0^{\pi}\frac{dx}{1+\sin^2x} = \int_0^{\pi}\frac{dx}{1+\cos^2x} = \frac{\pi}{\sqrt2}$ .
$\sin^4x+\cos^4x = \dfrac{3+\cos 4x}{4}$ , so $\int_0^{\pi}(\sin^4x+\cos^4x)\,dx=\tfrac{3\pi}{4}$ .
$1+\cos\theta = 2\cos^2\tfrac{\theta}{2}$ , $1-\cos\theta=2\sin^2\tfrac{\theta}{2}$ (half-angle).
Beta function: $\displaystyle\int_0^1 x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!}$ .

Standard definite-integral results

\int_0^{\pi}\frac{dx}{1+\sin^2x} = \frac{\pi}{\sqrt2} \qquad \int_0^1 x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!} \qquad \int_0^{\pi/2}\sin^{2}x\,dx = \int_0^{\pi/2}\cos^{2}x\,dx = \frac{\pi}{4}

Worked example

Evaluate

\displaystyle\int_0^1 x^2(1-x)^3\,dx

using the Beta result.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 3Definite IntegrationMODERATE

What is

\int_0^{\pi/2}\dfrac{d\theta}{1+\cos\theta}

equal to?

[Q72 · Apr · 2017]

Reduce the power BEFORE integrating

Don't integrate

\sin^4x

and

\cos^4x

separately — combine them via

\sin^4x+\cos^4x=\frac{3+\cos4x}{4}

first. The reduction turns a tedious computation into a one-line answer.

Drill 5 more on standard results and trig reductions

Concept 4 of 4

Direct evaluation — simplify, then integrate

Intuition

Many definite integrals are just antiderivative-and-substitute once you simplify the integrand: factor a trig expression, apply by-parts, or substitute. This concept is where the Indefinite Integration toolkit meets definite limits — the only new step is plugging in the bounds.

Definition

Simplification routes (then apply the limits):

Trig factoring: $\tan^3x+\tan x = \tan x\,\sec^2x$ , which integrates to $\tfrac{\tan^2x}{2}$ .
By-parts for products: $\int x\ln x\,dx$ , $\int e^x\sin x\,dx = \tfrac{e^x(\sin x-\cos x)}{2}$ .
Substitution: $\int_0^{\pi/2}e^{\sin x}\cos x\,dx$ via $u=\sin x$ .

(For the full substitution / by-parts technique, see the Indefinite Integration notes — here just carry the limits through.)

Worked example

Evaluate

\displaystyle\int_0^{\pi/2} e^{\sin x}\cos x\,dx

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 4Definite IntegrationMODERATE

What is the value of

\int_0^{\pi/4}(\tan^3 x+\tan x)\,dx

[Q79 · Apr · 2020]

Transform the limits when you substitute

When you substitute

u=g(x)

in a definite integral, change the limits to

g(a)

and

g(b)

and you never need to back-substitute. Forgetting to convert the limits is the classic definite-integral error.

Drill 4 more on direct evaluation — simplify, then integrate

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

King's property — the reflection trick
King's property and its companions
$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \qquad \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \qquad \int_0^{2a} f(x)\,dx = \int_0^a \big[f(x)+f(2a-x)\big]\,dx \qquad \int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}$
Odd/even symmetry over a symmetric interval
Odd/even symmetry over a symmetric interval
$\int_{-a}^{a} f(x)\,dx = 0 \;\;(f\text{ odd}) \qquad \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \;\;(f\text{ even}) \qquad \int_{-a}^{a} \frac{f(x)}{1+c^{x}}\,dx = \int_0^a f(x)\,dx \;\;(f\text{ even})$
Standard results and trig reductions
Standard definite-integral results
$\int_0^{\pi}\frac{dx}{1+\sin^2x} = \frac{\pi}{\sqrt2} \qquad \int_0^1 x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!} \qquad \int_0^{\pi/2}\sin^{2}x\,dx = \int_0^{\pi/2}\cos^{2}x\,dx = \frac{\pi}{4}$

Watch out for (5)

Add the reflected form — don't just substitute and stop
→ King's property — the reflection trick
King's reflection only works with the matching limits
→ King's property — the reflection trick
Check parity of the WHOLE integrand
→ Odd/even symmetry over a symmetric interval
Reduce the power BEFORE integrating
→ Standard results and trig reductions
Transform the limits when you substitute
→ Direct evaluation — simplify, then integrate

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