$\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x}$ is equal to (A) $1$ (B) $2$ (C) $3$ (D) $4$

∫-π/4π/4 (dx)/(1+cos 2x) is equal to (A) 1 (B) 2 (C) 3 (D) 4

NCERT Exemplar · Q62

\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x}

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Concept understanding — Definite Integral Symmetry

Definite Integral Symmetry: The Shortcut That Saves You Work

Asked to find the area under $f(x) = x^3$ from $x = -2$ to $x = 2$ ? You could integrate directly — but there's a much faster way if you notice the symmetry of the graph.

The Intuition: What Does "Symmetry" Mean Here?

A function can be symmetric about the y-axis (like $x^2$ or $\cos x$ ) or about the origin (like $x^3$ or $\sin x$ ). When you integrate over a symmetric interval — from $-a$ to $a$ — these symmetries create a predictable cancellation or doubling.

Even functions (symmetric about the y-axis): $f(-x) = f(x)$ — think $x^2$ , $x^4$ , $\cos x$ , $|x|$ . The left side mirrors the right, so the area from $-a$ to $0$ equals the area from $0$ to $a$ . The total is double the area on one side.

Odd functions (symmetric about the origin): $f(-x) = -f(x)$ — think $x^3$ , $x^5$ , $\sin x$ , $\tan x$ . The left side is the negative mirror of the right, so every positive area on the right is cancelled by an equal negative area on the left. The total is zero.

Note

This only works when the limits are symmetric about zero — from $-a$ to $a$ . For $[0, a]$ or $[1, 3]$ , symmetry doesn't help directly.

The Precise Statement

Let $f$ be continuous on $[-a, a]$ .

If $f$ is even ( $f(-x) = f(x)$ ), then $\;\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$
If $f$ is odd ( $f(-x) = -f(x)$ ), then $\;\int_{-a}^{a} f(x) \, dx = 0$

\int_{-a}^{a} f(x) \, dx = \begin{cases} 2 \int_{0}^{a} f(x) \, dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}

Why Does This Work? (A Quick Proof)

Split at zero:

\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx

For the first term substitute $u = -x$ ( $dx = -du$ ; $x=-a \to u=a$ , $x=0 \to u=0$ ):

\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-u) \, du

Now use symmetry:

If $f$ is even, $f(-u) = f(u)$ , so this becomes $\int_{0}^{a} f(u) \, du$ ; adding the second term gives $2 \int_{0}^{a} f(x) \, dx$ .
If $f$ is odd, $f(-u) = -f(u)$ , so this becomes $-\int_{0}^{a} f(u) \, du$ ; adding the second term gives $0$ .

Common Mistakes to Avoid

Watch out

Don't assume symmetry without checking. A "balanced"-looking function need not be even or odd — e.g. $f(x) = x^2 + x$ is neither, so these formulas don't apply. Always verify $f(-x) = f(x)$ or $f(-x) = -f(x)$ for all $x$ .

Watch out

The interval must be $[-a, a]$ . If the limits are $[-2, 3]$ , symmetry doesn't apply directly — split the integral or shift variables.

Examples to Cement the Idea

Example 1: $\int_{-3}^{3} x^4 \, dx$ — $x^4$ is even, so

$\int_{-3}^{3} x^4 \, dx = 2 \int_{0}^{3} x^4 \, dx = 2 \left[ \frac{x^5}{5} \right]_{0}^{3} = 2 \cdot \frac{243}{5} = \frac{486}{5}$ …

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The integral simplifies using the identity $\cos 2x = 2\cos^2 x - 1$ , turning the integrand into $\frac{1}{2}\sec^2 x$ . The definite integral from $-\pi/4$ to $\pi/4$ then evaluates to $1$ , so the correct option is (A).

The key here is to notice that the integrand looks messy, but a standard trigonometric identity will clean it up instantly. The denominator $1 + \cos 2x$ is a classic form that simplifies to $2\cos^2 x$ . Once you see that, the integral becomes a simple $\sec^2 x$ integral, which is the derivative of $\tan x$ . The symmetry of the limits $-\pi/4$ to $\pi/4$ also helps — the function becomes even, so you could double the integral from $0$ to $\pi/4$ , but it’s not necessary.

Let’s work through it step by step.

Simplify the denominator using the double-angle identity. Recall: $\cos 2x = 2\cos^2 x - 1$ . Therefore, $1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x$ . So the integral becomes:

\int_{-\pi/4}^{\pi/4} \frac{dx}{2\cos^2 x} = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx.

Integrate $\sec^2 x$ . The antiderivative of $\sec^2 x$ is $\tan x$ (since $\frac{d}{dx}\tan x = \sec^2 x$ ). So:

\frac{1}{2} \left[ \tan x \right]_{-\pi/4}^{\pi/4}.

Evaluate at the limits. $\tan(\pi/4) = 1$ , and $\tan(-\pi/4) = -1$ . Hence: $\frac{1}{2} \left( 1 - (-1) \right) = \frac{1}{2} \times 2 = 1.$ …

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Q. $\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x}$ is equal to
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Concept understanding — Definite Integral Symmetry

Definite Integral Symmetry: The Shortcut That Saves You Work

The Intuition: What Does "Symmetry" Mean Here?

The Precise Statement

Why Does This Work? (A Quick Proof)

Common Mistakes to Avoid

Examples to Cement the Idea

Method: Collapse $1\pm\cos 2x$ with a double-angle identity

Steps

Common Mistakes

Q.∫−π/4π/4dx1+cos⁡2x\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x}∫−π/4π/4​1+cos2xdx​ is equal to (A) 111 (B) 222 (C) 333 (D) 444

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Q. $\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x}$ is equal to
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Unlock everything free for 14 days