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.

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: 0af(x)dx=0af(ax)dx\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx (and bcf(x)dx=bcf(b+cx)dx\int_b^c f(x)\,dx=\int_b^c f(b+c-x)\,dx).
  • The 2I move: let I=0af(x)dxI=\int_0^a f(x)\,dx; write a second copy with xaxx\to a-x; ADD them so the denominators match. Often 2I=0a1dx=a2I=\int_0^a 1\,dx = a.
  • Classic family: 0af(x)f(x)+f(ax)dx=a2\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}.
  • Works on 0π/2sinxsinx+cosxdx=π4\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx = \frac{\pi}{4} and the whole acosx+bsinxsinx+cosx\frac{a\cos x+b\sin x}{\sin x+\cos x} family.

King's property

0af(x)dx=0af(ax)dx\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx
x = a/2f(x)f(a−x)0aequal areas → I = ∫f(x)dx = ∫f(a−x)dx

Worked example

Evaluate I=0π/2sinxsinx+cosxdxI=\displaystyle\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx.
  1. Apply xπ2xx\to \tfrac{\pi}{2}-x: sinxcosx\sin x\to\cos x, cosxsinx\cos x\to\sin x, so I=0π/2cosxcosx+sinxdxI=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}\,dx.
  2. Add the two forms: 2I=0π/2sinx+cosxsinx+cosxdx=0π/21dx=π22I=\int_0^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}\,dx=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.
  3. So I=π4I=\frac{\pi}{4}.
Answer:π4\frac{\pi}{4}.
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate I=0πxsinx1+cos2xdxI=\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    0af(ax)f(x)+f(ax)dx=?\int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\,dx = ?
  2. 2.
    0π/2cosxsinx+cosxdx=?\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx = ?
  3. 3.
    King's property: 0af(x)dx=?\int_0^a f(x)\,dx = ?
  4. 4.
    01ln ⁣(1xx)dx=?\int_0^1 \ln\!\big(\tfrac{1-x}{x}\big)dx = ?

From the bank · past-year question

Example 1Definite IntegrationMODERATE
What is 0π/2a+sinx2(a+sinx+cosx)dx\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 II; 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.

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][-a,a]:

  • Odd f(x)=f(x)f(-x)=-f(x): aaf(x)dx=0\displaystyle\int_{-a}^{a} f(x)\,dx = 0.
  • Even f(x)=f(x)f(-x)=f(x): aaf(x)dx=20af(x)dx\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^{a} f(x)\,dx.
  • **The 1+cx1+c^x property**: for even ff, aaf(x)1+cxdx=12aaf(x)dx=0af(x)dx\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 aah(x)dx=0a[h(x)+h(x)]dx\displaystyle\int_{-a}^{a} h(x)\,dx = \int_0^a [h(x)+h(-x)]\,dx.
ab∫ f dxy = f(x)

Worked example

Evaluate π/4π/4(sinxtanx)dx\displaystyle\int_{-\pi/4}^{\pi/4} (\sin x - \tan x)\,dx.
  1. Check parity: sin(x)=sinx\sin(-x)=-\sin x and tan(x)=tanx\tan(-x)=-\tan x, so the integrand is odd.
  2. An odd function over a symmetric interval integrates to 0.
Answer:0.
Practice this conceptself-check · 3 quick reps

Try it yourself

Evaluate π/2π/2(ecosxsinx+esinxcosx)dx\displaystyle\int_{-\pi/2}^{\pi/2}(e^{\cos x}\sin x + e^{\sin x}\cos x)\,dx.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    aa\int_{-a}^{a} of an odd function?
  2. 2.
    22x3cosxdx=?\int_{-2}^{2} x^3\cos x\,dx = ?
  3. 3.
    11x21+5xdx=?\int_{-1}^{1}\frac{x^2}{1+5^x}\,dx = ?

From the bank · past-year question

Example 2Definite IntegrationMODERATE
What is π/2π/2(ecosxsinx+esinxcosx)dx\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.

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:

  • 0πdx1+sin2x=0πdx1+cos2x=π2\displaystyle\int_0^{\pi}\frac{dx}{1+\sin^2x} = \int_0^{\pi}\frac{dx}{1+\cos^2x} = \frac{\pi}{\sqrt2}.
  • sin4x+cos4x=3+cos4x4\sin^4x+\cos^4x = \dfrac{3+\cos 4x}{4}, so 0π(sin4x+cos4x)dx=3π4\int_0^{\pi}(\sin^4x+\cos^4x)\,dx=\tfrac{3\pi}{4}.
  • 1+cosθ=2cos2θ21+\cos\theta = 2\cos^2\tfrac{\theta}{2}, 1cosθ=2sin2θ21-\cos\theta=2\sin^2\tfrac{\theta}{2} (half-angle).
  • Beta function: 01xm(1x)ndx=m!n!(m+n+1)!\displaystyle\int_0^1 x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!}.

Worked example

Evaluate 01x2(1x)3dx\displaystyle\int_0^1 x^2(1-x)^3\,dx using the Beta result.
  1. Here m=2, n=3m=2,\ n=3.
  2. Beta function: 2!3!(2+3+1)!=26720=12720\frac{2!\,3!}{(2+3+1)!} = \frac{2\cdot 6}{720} = \frac{12}{720}.
Answer:160\frac{1}{60}.
Practice this conceptself-check · 3 quick reps

Try it yourself

Evaluate 0π/2dθ1+cosθ\displaystyle\int_0^{\pi/2}\frac{d\theta}{1+\cos\theta}.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    0πdx1+sin2x=?\int_0^{\pi}\frac{dx}{1+\sin^2x} = ?
  2. 2.
    Write 1cosθ1-\cos\theta as a half-angle square.
  3. 3.
    01x(1x)9dx=?\int_0^1 x(1-x)^9\,dx = ?

From the bank · past-year question

Example 3Definite IntegrationMODERATE
What is 0π/2dθ1+cosθ\int_0^{\pi/2}\dfrac{d\theta}{1+\cos\theta} equal to?

[Q72 · Apr · 2017]

Reduce the power BEFORE integrating

Don't integrate sin4x\sin^4x and cos4x\cos^4x separately — combine them via sin4x+cos4x=3+cos4x4\sin^4x+\cos^4x=\frac{3+\cos4x}{4} first. The reduction turns a tedious computation into a one-line answer.

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: tan3x+tanx=tanxsec2x\tan^3x+\tan x = \tan x\,\sec^2x, which integrates to tan2x2\tfrac{\tan^2x}{2}.
  • By-parts for products: xlnxdx\int x\ln x\,dx, exsinxdx=ex(sinxcosx)2\int e^x\sin x\,dx = \tfrac{e^x(\sin x-\cos x)}{2}.
  • Substitution: 0π/2esinxcosxdx\int_0^{\pi/2}e^{\sin x}\cos x\,dx via u=sinxu=\sin x.

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

Worked example

Evaluate 0π/2esinxcosxdx\displaystyle\int_0^{\pi/2} e^{\sin x}\cos x\,dx.
  1. Substitute u=sinxu=\sin x, du=cosxdxdu=\cos x\,dx; limits x:0π2x:0\to\tfrac{\pi}{2} give u:01u:0\to 1.
  2. Integral =01eudu=[eu]01=e1=\int_0^1 e^u\,du = [e^u]_0^1 = e-1.
Answer:e1e-1.
Practice this conceptself-check · 3 quick reps

Try it yourself

Evaluate 0π/4(tan3x+tanx)dx\displaystyle\int_0^{\pi/4}(\tan^3x+\tan x)\,dx.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Simplify tan3x+tanx\tan^3x+\tan x.
  2. 2.
    exsinxdx=?\int e^x\sin x\,dx = ?
  3. 3.
    1exlnxdx=?\int_1^e x\ln x\,dx = ?

From the bank · past-year question

Example 4Definite IntegrationMODERATE
What is the value of 0π/4(tan3x+tanx)dx\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)u=g(x) in a definite integral, change the limits to g(a)g(a) and g(b)g(b) and you never need to back-substitute. Forgetting to convert the limits is the classic definite-integral error.

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 (1)

Watch out for (4)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Definite IntegrationMODERATE
What is 0πln ⁣(tanx2)dx\int_{0}^{\pi}\ln\!\left(\tan\frac{x}{2}\right)dx equal to?

[Q89 · Sep · 2021]

Example 2Definite IntegrationMODERATE
Let I=2π2πsin4x+cos4x1+3xdxI=\int_{-2\pi}^{2\pi}\frac{\sin^4x+\cos^4x}{1+3^x}\,dx.
What is 0π(sin4x+cos4x)dx\int_0^\pi(\sin^4x+\cos^4x)\,dx equal to?

[Q61 · Sep · 2023]

Example 3Definite IntegrationMODERATE
What is 01x(1x)9dx\int_{0}^{1}x(1-x)^9\,dx equal to?

[Q94 · Apr · 2018]

Example 4Definite IntegrationMODERATE
What is 1exlnxdx\int_{1}^{e}x\ln x\,dx equal to?

[Q81 · Apr · 2018]

Example 5Definite IntegrationMODERATE
What is 01ln ⁣(1x1)dx\int_0^1 \ln\!\left(\frac{1}{x}-1\right)dx equal to?

[Q94 · Sep · 2022]

Drill every past-year question on this subtopic

32 questions from the bank — paginated, with cart and Word-export support.