MHT-CET Maths · Definite Integration

Odd and Even Integrands — Symmetric Limits

Over limits symmetric about 0, an odd integrand integrates to 0 and an even one to twice the half — so the first thing to do with limits −a to a is test f(−x), before any antiderivative.

All Definite Integration notes MHT-CET Maths strategy

Why this matters

11 PYQs at 55% HARD, and the fastest marks in the chapter when the test is applied first: five of the eleven are answered by writing f(−x) = −f(x) and nothing else. The HARD ones hide the symmetry — a polynomial that becomes odd only after the variable is shifted, a log term that is odd while the rest is even, or a substitution x → 1/x that plays the role of x → −x. One recurring stem is a trap in the other direction: an even integrand whose integral does not exist at all, keyed to a formal value the exam accepts.

Concept 1 of 4

The Odd/Even Test on Symmetric Limits

Intuition

An odd function's graph on

[-a, 0]

is the upside-down mirror of its graph on

[0, a]

: the two signed areas cancel exactly. An even function's two halves are identical, so the whole is twice one half.

Definition

$f$ is even if $f(-x) = f(x)$ ; odd if $f(-x) = -f(x)$ . Test by replacing $x$ with $-x$ and simplifying.
$\int_{-a}^{a} f(x)\,dx = 0$ for odd $f$ ; $= 2\int_0^a f(x)\,dx$ for even $f$ .
Products: odd × odd = even, even × even = even, odd × even = odd. $\sin^7x\cos^{16}x$ is odd ( $\sin$ to an odd power); $x^2\cos x$ is even.
$f(x) + f(-x)$ is always even and $g(x) - g(-x)$ is always odd, whatever $f$ and $g$ are — so their product is odd and integrates to $0$ over any symmetric interval.
Symmetric limits that do not look symmetric: $\log\frac12$ to $\log 2$ is $-\log 2$ to $\log 2$ . And $\sin\left(\dfrac{e^x - 1}{e^x + 1}\right)$ is odd because $\dfrac{e^{-x} - 1}{e^{-x} + 1} = -\dfrac{e^x - 1}{e^x + 1}$ .

Symmetric-limit rule

\int_{-a}^{a} f(x)\,dx = \begin{cases} 0, & f(-x) = -f(x) \\[2pt] 2\displaystyle\int_0^a f(x)\,dx, & f(-x) = f(x) \end{cases}

Worked example

Evaluate

\int_{-2}^{2}\left(x^3\cos x + x^2\right)dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Definite IntegrationEASY

The value of

\int_{- 3}^{3} \sin^{7}x\cos^{16}x\text{ }dx

[Q134 · 19 April Shift II · 2025]

Testing the limits instead of the function

Symmetric limits are necessary, not sufficient.

\int_{-1}^{1}x^2\,dx

\frac23

, not

0

: the integrand must be odd for the integral to vanish. Write

f(-x)

explicitly every time.

Drill 3 more on the odd/even test on symmetric limits

Concept 2 of 4

Split a Mixed Integrand into Its Odd and Even Parts

Intuition

\dfrac{17x^5 - x^4 + 29x^3 - 31x + 1}{x^2 + 1}

looks hopeless, but every odd-power term over the even denominator is odd and vanishes on

[-1, 1]

; only

\dfrac{1 - x^4}{x^2 + 1} = 1 - x^2

survives. Split, discard, integrate what is left.

Definition

Any integrand splits as (odd part) + (even part). On $[-a, a]$ the odd part contributes $0$ ; integrate only the even part, doubled from $0$ to $a$ .
$\log\dfrac{\pi - x}{\pi + x}$ , $\log\dfrac{1 + x}{1 - x}$ and $\tan^{-1}x$ are odd; multiplied by an even function ( $\cos x$ ) they stay odd and drop out. What remains is typically $x^2\cos x$ , which needs by parts twice: $\int_0^{\pi/2}x^2\cos x\,dx = \dfrac{\pi^2}{4} - 2$ .
Shift the variable when the interval is symmetric about a point other than $0$ : on $[-2, 0]$ put $t = x + 1$ ; then $x^3 + 3x^2 + 3x + 5 = (x + 1)^3 + 4 = t^3 + 4$ and $(x + 1)\cos(x + 1) = t\cos t$ , both odd except the constant $4$ , so the integral is $4\times 2 = 8$ .
A greatest-integer term is neither odd nor even: $\int_{-1/2}^{1/2}[x]\,dx = \int_{-1/2}^{0}(-1)\,dx + 0 = -\dfrac12$ , evaluated by splitting, while the odd log beside it vanishes.

Odd part vanishes

\int_{-a}^{a}\big[\text{odd}(x) + \text{even}(x)\big]dx = 2\int_0^a \text{even}(x)\,dx \qquad \int_0^{\pi/2}x^2\cos x\,dx = \frac{\pi^2}{4} - 2

Worked example

Evaluate

\int_{-1}^{1}\dfrac{x^3 + x^2 + x + 1}{x^2 + 1}\,dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Definite IntegrationHARD

The value of

\int_{-1}^{1}\frac{17x^5 - x^4 + 29x^3 - 31x + 1}{x^2+1}\,dx

[Q123 · May Shift 1 · 2021]

Discarding the constant with the odd terms

t^3 + 4 + t\cos t

over

[-1, 1]

: the

t^3

and

t\cos t

vanish, but the

4

integrates to

8

. Zero is the answer for the ODD part, not for the question.

Drill 3 more on split a mixed integrand into its odd and even parts

Concept 3 of 4

Symmetry Under x → 1/x on [1/2, 2]

Intuition

On an interval like

\left[\frac12, 2\right]

, the substitution

x = \dfrac1t

maps the interval onto itself and flips the sign of

x - \dfrac1x

. If the integrand is

\dfrac1x

times an odd function of

x - \dfrac1x

, the integral equals its own negative — so it is

0

Definition

With $x = \dfrac1t$ : $dx = -\dfrac{dt}{t^2}$ , $\dfrac{dx}{x} = -\dfrac{dt}{t}$ , and $x - \dfrac1x = -\left(t - \dfrac1t\right)$ . Limits $\frac12 \to 2$ become $2 \to \frac12$ , and flipping them back cancels the minus from $dx$ .
So $I = \int_{1/2}^{2}\dfrac1x\,g\!\left(x - \dfrac1x\right)dx$ becomes $\int_{1/2}^{2}\dfrac1t\,g\!\left(-\left(t - \dfrac1t\right)\right)dt$ ; for odd $g$ this is $-I$ , hence $I = 0$ .
$\csc^{101}$ is odd (odd power of an odd function), so $\int_{1/2}^{2}\dfrac1x\csc^{101}\!\left(x - \dfrac1x\right)dx = 0$ .
The interval must be of the form $\left[\frac1a, a\right]$ and the $\dfrac1x$ factor must be present; without it the trick fails.

The reciprocal symmetry

x = \tfrac1t:\quad \int_{1/a}^{a}\frac{1}{x}\,g\!\left(x - \frac1x\right)dx = -\int_{1/a}^{a}\frac{1}{t}\,g\!\left(t - \frac1t\right)dt \ \text{ for odd } g \ \Rightarrow\ I = 0

Worked example

Evaluate

\int_{1/3}^{3}\dfrac{1}{x}\sin\!\left(x - \dfrac1x\right)dx

Practice this conceptself-check

From the bank · past-year question

Example 3Definite IntegrationHARD

\int_{\frac{1}{2}}^{2} \frac{1}{x}\csc^{101}\left( x-\frac{1}{x} \right)dx=

[Q142 · 26 April Shift I · 2025]

Applying it without the 1/x

\int_{1/2}^{2}\sin\!\left(x - \frac1x\right)dx

is NOT zero: without the

\dfrac1x

the substitution produces an extra

\dfrac{1}{t^2}

and the symmetry breaks. Check for the factor before claiming the cancellation.

Concept 4 of 4

Even Does Not Mean Convergent: the csc⁴x Trap

Intuition

\csc^4x

is even, and doubling the half-integral is the natural move — but

\csc^4x

blows up at

x = 0

, which sits inside

\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]

. The integral does not exist; what the exam key reports is the antiderivative evaluated end to end.

Definition

Before applying any property, check the integrand is defined and finite on the whole interval. $\csc x$ , $\sec x$ , $\tan x$ , $\dfrac1x$ and $\log x$ each have points where they blow up.
$\int_{-\pi/4}^{\pi/4}\csc^4x\,dx$ : the antiderivative is $-\cot x - \dfrac{\cot^3x}{3}$ ; evaluated formally from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ it gives $\left(-1 - \frac13\right) - \left(1 + \frac13\right) = -\dfrac83$ — a NEGATIVE number for a positive integrand, which is the tell that something is wrong. The true integral diverges.
This question has been set twice (2023 and 2025) with the official key $-\dfrac83$ ; on the paper, mark the key. In your understanding, know why it is meaningless.
The same care applies to $\int_0^{\pi/2}\dfrac{dx}{1 - \cos x}$ (blows up at $0$ ) and to any $\int\dfrac{dx}{x^2}$ across $0$ .

Check the domain first

\int_{-\pi/4}^{\pi/4}\csc^4x\,dx \text{ diverges}; \quad \left[-\cot x - \tfrac{\cot^3x}{3}\right]_{-\pi/4}^{\pi/4} = -\tfrac83 \text{ is the exam's formal key}

Worked example

Does

\int_{-1}^{1}\dfrac{dx}{x^2}

exist? What does the formal evaluation give?

Practice this concept4 quick reps

From the bank · past-year question

Example 4Definite IntegrationHARD

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

has the value

[Q131 · 10th May Shift 2 · 2023]

Marking the mathematically honest option

There is no honest option — the integral diverges and the key is

-\frac83

. On this paper choose the key; on any paper, a negative value for a positive integrand should make you re-read the interval for a blow-up.

Drill 1 more on even does not mean convergent: the csc⁴x trap

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 (4)

The Odd/Even Test on Symmetric Limits
Symmetric-limit rule
$\int_{-a}^{a} f(x)\,dx = \begin{cases} 0, & f(-x) = -f(x) \\[2pt] 2\displaystyle\int_0^a f(x)\,dx, & f(-x) = f(x) \end{cases}$
Split a Mixed Integrand into Its Odd and Even Parts
Odd part vanishes
$\int_{-a}^{a}\big[\text{odd}(x) + \text{even}(x)\big]dx = 2\int_0^a \text{even}(x)\,dx \qquad \int_0^{\pi/2}x^2\cos x\,dx = \frac{\pi^2}{4} - 2$
Symmetry Under x → 1/x on [1/2, 2]
The reciprocal symmetry
$x = \tfrac1t:\quad \int_{1/a}^{a}\frac{1}{x}\,g\!\left(x - \frac1x\right)dx = -\int_{1/a}^{a}\frac{1}{t}\,g\!\left(t - \frac1t\right)dt \ \text{ for odd } g \ \Rightarrow\ I = 0$
Even Does Not Mean Convergent: the csc⁴x Trap
Check the domain first
$\int_{-\pi/4}^{\pi/4}\csc^4x\,dx \text{ diverges}; \quad \left[-\cot x - \tfrac{\cot^3x}{3}\right]_{-\pi/4}^{\pi/4} = -\tfrac83 \text{ is the exam's formal key}$

Watch out for (4)

Testing the limits instead of the function
→ The Odd/Even Test on Symmetric Limits
Discarding the constant with the odd terms
→ Split a Mixed Integrand into Its Odd and Even Parts
Applying it without the 1/x
→ Symmetry Under x → 1/x on [1/2, 2]
Marking the mathematically honest option
→ Even Does Not Mean Convergent: the csc⁴x Trap

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