MHT-CET Maths · Definite Integration

King's Property — f(a + b − x) and the f/(f + g) Family

Replacing x by a + b − x leaves a definite integral unchanged — and adding the two forms cancels the awkward part, turning an unintegrable-looking expression into a constant times the interval.

All Definite Integration notes MHT-CET Maths strategy

Why this matters

28 PYQs at 50% HARD, the largest property page in the chapter and the highest-leverage recognition in MHT-CET calculus: in the 2021–2025 papers, eleven of the seventeen were answered by writing the reflected integral, adding, and dividing by two. Five distinct families recur — f/(f + g) over an interval, x·f(sin x) over 0 to π, integrands with 1/(1 + eˣ), functional equations like f(x) = f(1 − x), and an inverse-trig identity applied before the reflection — and each has a one-line closed form worth knowing. The tell is always the same: the interval's endpoints add to something that makes the reflected integrand look like the original.

Concept 1 of 6: King's Property: ∫ f(x) = ∫ f(a + b − x)

Reflecting the graph across the midpoint of

[a, b]

does not change the area under it. So

\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx

— and if the reflected integrand is simpler, or combines nicely with the original, the question is over.

Definition

$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ ; in particular $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ .
Method: write $I$ with $x \to a + b - x$ , call it the same $I$ , ADD the two expressions, simplify the sum, and divide by $2$ .
$\int_0^{\pi/4}\log(1 + \tan x)\,dx$ : with $x \to \frac{\pi}{4} - x$ , $1 + \tan\left(\frac{\pi}{4} - x\right) = \dfrac{2}{1 + \tan x}$ , so $2I = \int_0^{\pi/4}\log 2\,dx = \dfrac{\pi}{4}\log 2$ and $I = \dfrac{\pi}{8}\log 2$ . The same question is set as $\log\dfrac{\sin x + \cos x}{\cos x}$ .
Handy reflections: on $[0, \frac{\pi}{2}]$ , $\sin \leftrightarrow \cos$ , $\tan \leftrightarrow \cot$ ; on $[0, \pi]$ , $\sin x$ is unchanged and $\cos x \to -\cos x$ ; on $[a, b]$ with $a + b = 8$ , $\sqrt{x} \leftrightarrow \sqrt{8 - x}$ .

King's property

\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx \qquad \int_0^{\pi/4}\log(1 + \tan x)\,dx = \frac{\pi}{8}\log 2

Worked example

Evaluate

\int_0^{\pi/2}\log\!\left(\dfrac{\sin x}{\sin x + \cos x}\right)dx

given that

\int_0^{\pi/2}\log\sin x\,dx = -\dfrac{\pi}{2}\log 2

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q115Moderate

Example 1 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

I = \int_0^{\pi/4}\log(1+\tan x)\,dx

, then value of I is

Reflecting about the wrong point

[\pi/3, 2\pi/3]

the reflection is

x \to \pi - x

(endpoints add to

\pi

), not

x \to \frac{\pi}{2} - x

. Always add the two endpoints first; that sum is the only thing the substitution uses.

Drill 1 more on king's property: ∫ f(x) = ∫ f(a + b − x)

Concept 2 of 6: The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2

If the denominator is the numerator plus its own reflection, then adding

I

to its reflected copy gives

\int 1\,dx = b - a

. Half of that is the answer — no integration performed at all.

Definition

$\int_a^b\dfrac{f(x)}{f(x) + f(a + b - x)}\,dx = \dfrac{b - a}{2}$ .
On $[0, \frac{\pi}{2}]$ : $\dfrac{\cot^n x}{\cot^n x + \tan^n x}$ , $\dfrac{1}{1 + \cot^{101}x}$ , $\dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}$ all give $\dfrac{\pi}{4}$ .
On $[3, 5]$ : $\dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{8 - x}} \to 1$ . On $[2, 4]$ : $\dfrac{\log x^2}{\log x^2 + \log(6 - x)^2} \to 1$ , after factoring $36 - 12x + x^2 = (6 - x)^2$ ; on $[1, 3]$ with $16x^2 - 8x^3 + x^4 = x^2(4 - x)^2$ likewise.
Weighted numerators: $\int_0^{\pi/2}\dfrac{a\sin x + b\cos x}{\sin x + \cos x}\,dx = \dfrac{\pi}{4}(a + b)$ — reflect, add, the numerators sum to $(a + b)(\sin x + \cos x)$ .
The $\sin(x^2)$ version on $[\sqrt{\log 2}, \sqrt{\log 3}]$ : substitute $t = x^2$ first so the limits become $\log 2, \log 3$ with sum $\log 6$ , then the family applies with an extra $\dfrac12$ from $dt = 2x\,dx$ .

The f/(f + g) result

\int_a^b\frac{f(x)}{f(x) + f(a + b - x)}\,dx = \frac{b - a}{2} \qquad \int_0^{\pi/2}\frac{a\sin x + b\cos x}{\sin x + \cos x}\,dx = \frac{\pi}{4}(a + b)

Worked example

Evaluate

\int_1^4\dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{5 - x}}\,dx

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q122Moderate

Example 2 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

\int_{3}^{5} \frac{\sqrt{x}\text{ }dx}{\sqrt{8 -x}+\sqrt{x}}=

Missing the disguised reflection in the denominator

\log(16x^2 - 8x^3 + x^4) = \log x^2 + \log(4 - x)^2

: on

[1, 3]

the second term IS the reflection of the first (

1 + 3 = 4

). Factor the quartic before deciding the family does not apply.

Drill 6 more on the f/(f + g) family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2

Concept 3 of 6: The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2

Reflecting

x \to \pi - x

leaves

\sin x

alone and turns

x

into

\pi - x

. Adding the two copies replaces

x

\pi

, so the

x

factor comes out as

\dfrac{\pi}{2}

and an integral you can do is left.

Definition

$\int_0^\pi x\,f(\sin x)\,dx = \dfrac{\pi}{2}\int_0^\pi f(\sin x)\,dx$ . More generally, on any interval where $f(a + b - x) = f(x)$ , $\int_a^b x f(x)\,dx = \dfrac{a + b}{2}\int_a^b f(x)\,dx$ .
$\dfrac{x\tan x}{\sec x + \cos x} = \dfrac{x\sin x}{1 + \cos^2x}$ : the remaining $\int_0^\pi\dfrac{\sin x}{1 + \cos^2x}\,dx = \left[-\tan^{-1}(\cos x)\right]_0^\pi = \dfrac{\pi}{2}$ , so the total is $\dfrac{\pi}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi^2}{4}$ .
$\int_{\pi/3}^{2\pi/3}\dfrac{x}{1 + \sin x}\,dx$ : endpoints add to $\pi$ and $\sin(\pi - x) = \sin x$ , so $2I = \pi\int_{\pi/3}^{2\pi/3}\dfrac{dx}{1 + \sin x} = \pi[\tan x - \sec x]_{\pi/3}^{2\pi/3} = 2\pi(2 - \sqrt3)$ .
$\dfrac{1}{1 + \sin x} = \dfrac{1 - \sin x}{\cos^2x} = \sec^2x - \sec x\tan x$ is the standard way to integrate that piece.

Pulling x out

\int_0^\pi x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx \qquad \int_0^\pi\frac{\sin x}{1 + \cos^2x}\,dx = \frac{\pi}{2}

Worked example

Evaluate

\int_0^\pi x\sin x\,dx

using the reflection.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q106Hard

Example 3 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

\displaystyle\int_0^\pi \frac{x\tan x}{\sec x+\cos x}\,dx =

Using the trick with cos x

\cos(\pi - x) = -\cos x

, so

\int_0^\pi x\,g(\cos x)\,dx

does NOT reduce this way unless

g

is even. The trick needs the rest of the integrand to be unchanged by the reflection.

Drill 1 more on the x·f(sin x) trick on [0, π]: pull the x out as π/2

Concept 4 of 6: Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4

When the stem hands you

f(x) = f(1 - x)

, it is telling you the reflection leaves

f

alone. Reflect the integral, add, and the

x

g

factor collapses to a constant.

Definition

$f(x) = f(1 - x)$ on $[-1, 2]$ (endpoints add to $1$ ): $R_1 = \int_{-1}^{2}x f(x)\,dx = \int_{-1}^{2}(1 - x)f(x)\,dx$ , so $2R_1 = \int_{-1}^{2}f(x)\,dx = R_2$ and $R_2 = 2R_1$ .
$I_1 = \int_{1-h}^{h}x f(x(1 - x))\,dx$ , $I_2$ the same without $x$ : the reflection $x \to 1 - x$ fixes $x(1 - x)$ , so $I_1 = I_2 - I_1$ and $\dfrac{I_1}{I_2} = \dfrac12$ .
$f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$ : $I = \int_0^a f g = \int_0^a f(4 - g)$ , so $2I = 4\int_0^a f$ and $I = 2\int_0^a f$ .
Read the given identity as 'the reflection is free'; the rest is the add-and-halve routine.

Symmetry handed to you

f(a + b - x) = f(x) \ \Rightarrow\ \int_a^b x f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx

Worked example

f(x) = f(4 - x)

for all

x

and

\int_0^4 f(x)\,dx = 6

, find

\int_0^4 x f(x)\,dx

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q136Hard

Example 4 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

Let

f:[-1,2]\to[0,\infty)

be a continuous function such that

f(x)=f(1-x),\;\forall x\in[-1,2]

. Let

R_1=\int_{-1}^{2}xf(x)\,dx

and

R_2

be the area of the region bounded by

y=f(x)

x=-1

x=2

and the X-axis, then

R_2

Treating R₂ as an integral of x f(x)

R_2

is the AREA under

f

, i.e.

\int f

, with no

x

. The relation

R_2 = 2R_1

comes from

2R_1 = \int f

; reading it the other way round gives

\frac12 R_1

, which is offered.

Drill 2 more on functional symmetry given in the stem: f(x) = f(1 − x), g(x) + g(a − x) = 4

Concept 5 of 6: Integrands with 1/(1 + aˣ) over Symmetric Limits

\dfrac{1}{1 + e^{x}}

and

\dfrac{1}{1 + e^{-x}}

add up to exactly

1

. So over

[-a, a]

, reflecting

x \to -x

and adding wipes the exponential out, leaving the even factor integrated over half the interval.

Definition

$\dfrac{1}{1 + a^{x}} + \dfrac{1}{1 + a^{-x}} = 1$ for any base $a$ ( $e$ , $2$ , anything).
For even $f$ : $\int_{-a}^{a}\dfrac{f(x)}{1 + e^{x}}\,dx = \int_0^a f(x)\,dx$ . Reflect $x \to -x$ , add: $2I = \int_{-a}^{a}f = 2\int_0^a f$ .
$\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1 + 2^{x}}\,dx = \int_0^{\pi/2}\sin^2x\,dx = \dfrac{\pi}{4}$ .
$\int_{-\pi/2}^{\pi/2}\dfrac{x^2\cos x}{1 + e^{-x}}\,dx = \int_0^{\pi/2}x^2\cos x\,dx = \dfrac{\pi^2}{4} - 2$ (by parts twice for the last step).

The 1/(1 + aˣ) cancellation

\frac{1}{1 + a^{x}} + \frac{1}{1 + a^{-x}} = 1 \qquad f \text{ even}:\ \int_{-a}^{a}\frac{f(x)}{1 + e^{x}}\,dx = \int_0^a f(x)\,dx

Worked example

Evaluate

\int_{-1}^{1}\dfrac{x^2}{1 + e^{x}}\,dx

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q122Moderate

Example 5 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

The value of

\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x}\,dx

Forgetting the halving

After adding,

2I = \int_{-a}^{a}f

, which for even

f

2\int_0^a f

— so

I = \int_0^a f

, not

2\int_0^a f

. The option

\frac{\pi}{2}

beside the correct

\frac{\pi}{4}

is this slip.

Drill 1 more on integrands with 1/(1 + aˣ) over symmetric limits

Concept 6 of 6: An Inverse-Trig Identity Before the Reflection

\tan^{-1}(1 - x + x^2)

hides

\tan^{-1}x + \tan^{-1}(1 - x)

inside a

\cot^{-1}

; once unpacked, the two arctangents are reflections of each other on

[0, 1]

and integrate to the same thing.

Definition

$\tan^{-1}u = \dfrac{\pi}{2} - \cot^{-1}u$ , and $\cot^{-1}(1 - x + x^2) = \tan^{-1}\dfrac{1}{1 - x + x^2} = \tan^{-1}\dfrac{x + (1 - x)}{1 - x(1 - x)} = \tan^{-1}x + \tan^{-1}(1 - x)$ .
So $\int_0^1\tan^{-1}(1 - x + x^2)\,dx = \dfrac{\pi}{2} - \int_0^1\tan^{-1}x\,dx - \int_0^1\tan^{-1}(1 - x)\,dx = \dfrac{\pi}{2} - 2\int_0^1\tan^{-1}x\,dx$ (the last two are equal by King's property).
With $\int_0^1\tan^{-1}x\,dx = \dfrac{\pi}{4} - \dfrac12\log 2$ : the result is $\dfrac{\pi}{2} - \dfrac{\pi}{2} + \log 2 = \log 2$ .
The addition formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\dfrac{A + B}{1 - AB}$ (for $AB < 1$ ) is the tool; look for a quadratic argument that factors as $1 - AB$ with $A + B$ upstairs.

Unpack, then reflect

\cot^{-1}(1 - x + x^2) = \tan^{-1}x + \tan^{-1}(1 - x) \qquad \int_0^1\tan^{-1}(1 - x + x^2)\,dx = \log 2

Worked example

Evaluate

\int_0^1\cot^{-1}(1 - x + x^2)\,dx

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q121Hard

Example 6 · Definite Integration · King's Property — f(a + b − x) and the f/(f + g) Family

The value of

\int_{0}^{1} \tan^{- 1}\left( 1 -x+x^{2} \right)dx

Integrating tan⁻¹(1 − x + x²) directly

By parts on the quadratic argument produces a rational integral that takes minutes. The identity is the intended route and reduces the question to a known value of

\int_0^1\tan^{-1}x\,dx

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 (6)

King's Property: ∫ f(x) = ∫ f(a + b − x)
King's property
$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx \qquad \int_0^{\pi/4}\log(1 + \tan x)\,dx = \frac{\pi}{8}\log 2$
The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2
The f/(f + g) result
$\int_a^b\frac{f(x)}{f(x) + f(a + b - x)}\,dx = \frac{b - a}{2} \qquad \int_0^{\pi/2}\frac{a\sin x + b\cos x}{\sin x + \cos x}\,dx = \frac{\pi}{4}(a + b)$
The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2
Pulling x out
$\int_0^\pi x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx \qquad \int_0^\pi\frac{\sin x}{1 + \cos^2x}\,dx = \frac{\pi}{2}$
Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4
Symmetry handed to you
$f(a + b - x) = f(x) \ \Rightarrow\ \int_a^b x f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx$
Integrands with 1/(1 + aˣ) over Symmetric Limits
The 1/(1 + aˣ) cancellation
$\frac{1}{1 + a^{x}} + \frac{1}{1 + a^{-x}} = 1 \qquad f \text{ even}:\ \int_{-a}^{a}\frac{f(x)}{1 + e^{x}}\,dx = \int_0^a f(x)\,dx$
An Inverse-Trig Identity Before the Reflection
Unpack, then reflect
$\cot^{-1}(1 - x + x^2) = \tan^{-1}x + \tan^{-1}(1 - x) \qquad \int_0^1\tan^{-1}(1 - x + x^2)\,dx = \log 2$

Watch out for (6)

Reflecting about the wrong point
→ King's Property: ∫ f(x) = ∫ f(a + b − x)
Missing the disguised reflection in the denominator
→ The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2
Using the trick with cos x
→ The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2
Treating R₂ as an integral of x f(x)
→ Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4
Forgetting the halving
→ Integrands with 1/(1 + aˣ) over Symmetric Limits
Integrating tan⁻¹(1 − x + x²) directly
→ An Inverse-Trig Identity Before the Reflection

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