Board 2026 Class 12 Maths Q28 Solution: Show that I₁ − 4I₂ = 0 for two definite integrals

The question

If $I_1 = \displaystyle\int_{-\pi/4}^{\pi/4}\frac{dx}{1 + \cos 2x}$ and $I_2 = \displaystyle\int_{-1/2}^{1/2}|x|\,dx$ ⁠, then show that $I_1 - 4I_2 = 0$ ⁠.

Exam answer

Let $f(x) = \frac{1}{1 + \cos 2x}$ ⁠. Then $f(-x) = \frac{1}{1 + \cos(-2x)}$ $= \frac{1}{1 + \cos 2x} = f(x)$ ⁠, so $f$ is an even function.

Why?

Both integrals have limits of the form $-a$ to $a$ ⁠. For such limits, first test whether the integrand is even or odd.

$\cos$ is an even function, $\cos(-\theta) = \cos\theta$ ⁠, so $\cos(-2x) = \cos 2x$ ⁠.

Using $\displaystyle\int_{-a}^{a}f(x)\,dx = 2\int_0^a f(x)\,dx$ for an even function,

Why?

The graph of an even function is symmetric about the $y$ -axis, so the part from $-a$ to 0 equals the part from 0 to $a$ ⁠. That is why the integral doubles.

The new lower limit 0 makes the evaluation easier.

\begin{aligned} I_1 &= 2\int_0^{\pi/4}\frac{dx}{1 + \cos 2x} \\ &= 2\int_0^{\pi/4}\frac{dx}{2\cos^2x} \\ &= \int_0^{\pi/4}\sec^2x\,dx \\ &= \Big[\tan x\Big]_0^{\pi/4} = 1 - 0 = 1 \end{aligned}

Why?

The double-angle identity removes the 1:

1 + \cos 2x = 2\cos^2x

The 2 cancels, and $\frac{1}{\cos^2x} = \sec^2x$ ⁠.

\begin{aligned} 2\cdot\frac{1}{2\cos^2x} &= \frac{2}{2\cos^2x} \\ &= \frac{1}{\cos^2x} = \sec^2x \end{aligned}

Standard integral: $\int\sec^2x\,dx = \tan x$ ⁠. Then $\tan\frac{\pi}{4} = 1$ and $\tan 0 = 0$ ⁠.

\begin{aligned} \Big[\tan x\Big]_0^{\pi/4} &= \tan\frac{\pi}{4} - \tan 0 \\ &= 1 - 0 = 1 \end{aligned}

Also $|-x| = |x|$ ⁠, so $|x|$ is an even function, and $|x| = x$ for $x \ge 0$ ⁠.

Why?

The modulus must be removed before integrating. On $\left[-\frac12, \frac12\right]$ the sign of $x$ changes, so $|x|$ would need two pieces. The even-function property moves the work to $\left[0, \frac12\right]$ ⁠, where $x \ge 0$ and $|x| = x$ ⁠.

\begin{aligned} I_2 &= 2\int_0^{1/2}|x|\,dx = 2\int_0^{1/2}x\,dx \\ &= \Big[x^2\Big]_0^{1/2} = \frac14 \end{aligned}

Why?

$\int x\,dx = \frac{x^2}{2}$ ⁠, and the 2 in front cancels the $\frac12$ ⁠, leaving $x^2$ ⁠.

$\left(\frac12\right)^2 - 0^2 = \frac14$ ⁠.

\begin{aligned} 2\int_0^{1/2}x\,dx &= 2\left[\frac{x^2}{2}\right]_0^{1/2} \\ &= \Big[x^2\Big]_0^{1/2} \\ &= \left(\tfrac12\right)^2 - 0^2 \\ &= \tfrac14 - 0 = \tfrac14 \end{aligned}

\therefore\ I_1 - 4I_2 = 1 - 4\cdot\frac14 = 1 - 1

Why?

Put $I_1 = 1$ and $I_2 = \frac14$ ⁠:

\begin{aligned} 4\cdot\frac14 &= \frac44 = 1 \\ 1 - 1 &= 0 \end{aligned}

$\therefore\ I_1 - 4I_2 = 0$ ⁠. Hence proved.

Where this comes from

Chapter 7, Integrals: Definite integrals: even-function property.

Question type: Limits from $-a$ to $a$

How to spot it: click to see

Type 11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: Both pairs of limits are opposites, $\pm\frac{\pi}{4}$ and $\pm\frac12$ ⁠, and $\cos(-2x) = \cos 2x$ ⁠, $|-x| = |x|$ ⁠: both integrands are even. Double each integral from 0, where $|x| = x$ ⁠, to get $I_1 = 1$ and $I_2 = \frac14$ ⁠.

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Q28: Show that I1−4I2=0I_1 - 4I_2 = 0I1​−4I2​=0 for two definite integrals

Q28: Show that $I_1 - 4I_2 = 0$ for two definite integrals