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JEE Mains Maths · Definite Integration

The a + b − x Property and Other Symmetries

Writing a definite integral a second way, with x replaced by a + b − x or by 1/x, and adding the two forms so that the hard part cancels.

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Why this matters

Twenty-seven PYQs, and nearly all look impossible to integrate directly. They are built so that the integral written a second way, with the variable reflected, adds to the first to give something simple. Two ideas cover the page.

Concept 1 of 2: The a + b − x property

Reflecting the interval [a,b][a,b] in its midpoint does not change the integral: ∫abf(x) dx=∫abf(a+b−x) dx\int_a^bf(x)\,dx=\int_a^bf(a+b-x)\,dx. Write the integral both ways and add. The sum is often simple: sin⁡nxsin⁡nx+cos⁡nx\frac{\sin^nx}{\sin^nx+\cos^nx} and its reflection add to 1 on [0,π2]\left[0,\frac\pi2\right], and in ∫0πxf(sin⁡x) dx\int_0^\pi xf(\sin x)\,dx the xx and π−x\pi-x add to π\pi.

Definition

  • ∫abf(x) dx=∫abf(a+b−x) dx\int_a^bf(x)\,dx=\int_a^bf(a+b-x)\,dx.
  • ∫0π/2sin⁡nxsin⁡nx+cos⁡nxdx=π4\int_0^{\pi/2}\frac{\sin^nx}{\sin^nx+\cos^nx}dx=\frac\pi4 for every nn.
  • ∫0πxf(sin⁡x) dx=π2∫0πf(sin⁡x) dx\int_0^\pi xf(\sin x)\,dx=\frac\pi2\int_0^\pi f(\sin x)\,dx.
  • f(x)+f(a+b−x)=c⇒∫abf=c2(b−a)f(x)+f(a+b-x)=c\Rightarrow\int_a^bf=\frac c2(b-a).

Reflection in the midpoint

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

Worked example

Evaluate ∫0π/2sin⁡3xsin⁡3x+cos⁡3xdx\int_0^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q134Moderate

Example 1 · Definite Integration · The a + b - x Property and Other Symmetries

The integral ∫0π8xdx4cos⁡2x+sin⁡2x\int_{0}^{\pi} \frac{8xdx}{4\cos^{2}x+\sin^{2}x} is equal to.

The sum must be simpler

The property always holds, but it helps only when the two forms add to something easier. If they do not, the question needs a different method.
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Concept 2 of 2: The x → 1/x substitution

On limits 1a\frac1a to aa, or 0 to ∞\infty, putting x=1tx=\frac1t maps the interval onto itself, reversed, with dx=−dtt2dx=-\frac{dt}{t^2}. Adding the two forms often cancels a logarithm, or uses tan⁡−1x+tan⁡−11x=π2\tan^{-1}x+\tan^{-1}\frac1x=\frac\pi2. For f(x)=∫1xg(t) dtf(x)=\int_1^x g(t)\,dt, the sum f(x)+f(1x)f(x)+f\left(\frac1x\right) combines two integrands over the same range.

Definition

  • x=1tx=\frac1t: dx=−dtt2dx=-\frac{dt}{t^2}; [1a,a]\left[\frac1a,a\right] maps to itself.
  • ∫0∞ln⁡x1+x2dx=0\int_0^\infty\frac{\ln x}{1+x^2}dx=0.
  • tan⁡−1x+tan⁡−11x=π2\tan^{-1}x+\tan^{-1}\frac1x=\frac\pi2 for x>0x>0.

Reciprocal substitution

∫1/aaf(x) dx=∫1/aaf ⁣(1x)dxx2\int_{1/a}^{a} f(x)\,dx=\int_{1/a}^{a} f\!\left(\tfrac1x\right)\frac{dx}{x^2}

Worked example

Evaluate ∫0∞ln⁡x1+x2dx\int_0^\infty\frac{\ln x}{1+x^2}dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q159Moderate

Example 2 · Definite Integration · The a + b - x Property and Other Symmetries

The value of the integral ∫1/22tan⁡−1xxdx\int_{1/2}^{2} \frac{\tan^{- 1}x}{x}dx is equal to

Only for positive x

tan⁡−1x+tan⁡−11x=π2\tan^{-1}x+\tan^{-1}\frac1x=\frac\pi2 holds for x>0x>0; for x<0x<0 the sum is −π2-\frac\pi2.
Drill 3 more on the x → 1/x substitution

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 (2)

  • The a + b − x property

    Reflection in the midpoint

    ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx
  • The x → 1/x substitution

    Reciprocal substitution

    ∫1/aaf(x) dx=∫1/aaf ⁣(1x)dxx2\int_{1/a}^{a} f(x)\,dx=\int_{1/a}^{a} f\!\left(\tfrac1x\right)\frac{dx}{x^2}

Watch out for (2)

Test yourself on Definite Integration

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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