Evaluate: $\int_{0}^{\pi} x\sin x\cos^2 x\,dx$

Evaluate: ∫₀π xsin xcos² xdx

NCERT Exemplar · Q33

\int_{0}^{\pi} x\sin x\cos^2 x\,dx

Puducherry CbseShort· 3mImportance★★★★★

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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 symmetry property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ simplifies the integral to a standard form. The value is $\boxed{\frac{\pi}{3}}$ .

Why symmetry works here

When you see an integral from $0$ to $\pi$ with a product of $x$ and a trigonometric function, your first instinct should be: can I use the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ ? This is one of the most powerful tricks for definite integrals where the integrand involves $x$ multiplied by something symmetric.

The key insight: the factor $x$ is "antisymmetric" about $\pi/2$ (it increases from $0$ to $\pi$ ), while $\sin x \cos^2 x$ is symmetric about $\pi/2$ (check: $\sin(\pi-x)=\sin x$ , $\cos(\pi-x)=-\cos x$ , so $\cos^2$ is unchanged). When you replace $x$ by $\pi-x$ , the $x$ becomes $\pi-x$ , and the trigonometric part stays the same. Adding the original and transformed integrals lets you replace the troublesome $x$ with a constant $\pi/2$ .

Let's work through it.

Set up the transformation

Let $I = \int_{0}^{\pi} x \sin x \cos^2 x \, dx$ .

Use the property: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ . Here $a=\pi$ , so replace $x$ by $\pi-x$ :

I = \int_{0}^{\pi} (\pi - x) \sin(\pi - x) \cos^2(\pi - x) \, dx

Simplify the trigonometric terms

Recall: $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$ , so $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$ .

Therefore:

I = \int_{0}^{\pi} (\pi - x) \sin x \cos^2 x \, dx

Add the two expressions for $I$

We now have two forms of $I$ :

I = \int_{0}^{\pi} x \sin x \cos^2 x \, dx

I = \int_{0}^{\pi} (\pi - x) \sin x \cos^2 x \, dx

Add them:

2I = \int_{0}^{\pi} \big[x + (\pi - x)\big] \sin x \cos^2 x \, dx = \int_{0}^{\pi} \pi \sin x \cos^2 x \, dx

The $x$ terms cancel beautifully, leaving only a constant $\pi$ times the trigonometric part.

2I = \pi \int_{0}^{\pi} \sin x \cos^2 x \, dx

Evaluate the remaining integral

Let $J = \int_{0}^{\pi} \sin x \cos^2 x \, dx$ . …

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Q.Evaluate: $\int_{0}^{\pi} x\sin x\cos^2 x\,dx$

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

Why symmetry works here

Method: King Property for $\int_0^{\pi} x\,f(\sin x,\cos^2 x)\,dx$

Steps

Common Mistakes

Q.Evaluate: ∫0πxsin⁡xcos⁡2x dx\int_{0}^{\pi} x\sin x\cos^2 x\,dx∫0π​xsinxcos2xdx

Why symmetry works here

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Q.Evaluate: $\int_{0}^{\pi} x\sin x\cos^2 x\,dx$

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