The value of $\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx$ is _______.

The value of ∫-ππ sin 3xcos 2xdx is ____.

NCERT Exemplar · Q52

\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx

Yanam BieapShort· 1mImportance★★★★★

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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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Using the property that the integral of an odd function over a symmetric interval $[-a, a]$ is zero, we identify the integrand $\sin 3x \cos 2x$ as an odd function. Hence the integral evaluates to $0$ .

The key to solving this integral lies in recognising symmetry — specifically, whether the function you're integrating is odd or even over the interval $[-\pi, \pi]$ .

A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. For an odd function, the definite integral over a symmetric interval $[-a, a]$ is always zero:

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

This is because the area on the left side of the y-axis exactly cancels the area on the right side.

Now, look at our integrand: $\sin 3x \cdot \cos 2x$ . We need to check if it's odd.

Check the parity of each factor.
- $\sin 3x$ is an odd function: $\sin(-3x) = -\sin 3x$ .
- $\cos 2x$ is an even function: $\cos(-2x) = \cos 2x$ .
Combine them.

The product of an odd function and an even function is odd. Let's verify:

Let $f(x) = \sin 3x \cos 2x$ . Then

f(-x) = \sin(-3x) \cos(-2x) = (-\sin 3x)(\cos 2x) = -\sin 3x \cos 2x = -f(x)

So indeed $f(x)$ is odd.

Apply the symmetry property. Since the interval $[-\pi, \pi]$ is symmetric about zero and $f(x)$ is odd, we have: …

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Q.The value of $\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx$ is _______.

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: Parity test before integrating over $[-a,a]$

Steps

Common Mistakes

Q.The value of ∫−ππsin⁡3x cos⁡2x dx\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx∫−ππ​sin3xcos2xdx is _______.

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Q.The value of $\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx$ is _______.

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