Evaluate $\int_{-1}^1 \sin^5 x \cos^4 x\, dx$

Evaluate ∫-1¹ sin⁵ x cos⁴ x dx

Worked Examples · Example 31

\int_{-1}^1 \sin^5 x \cos^4 x\, dx

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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 integrand $\sin^5 x \cos^4 x$ is an odd function over a symmetric interval $[-1, 1]$ , so the integral is zero. The value is $\boxed{0}$ .

Why This Problem Is About Symmetry, Not Computation

If you try to compute $\int_{-1}^1 \sin^5 x \cos^4 x \, dx$ by expanding or using substitution, you’ll end up with a messy trigonometric integral. But there’s a much cleaner path: look at the function’s symmetry.

The interval $[-1, 1]$ is symmetric about $0$ . For such intervals, the integral of an odd function is always zero — provided the integral converges (which it does here, since the integrand is continuous). So the real question is: is $f(x) = \sin^5 x \cos^4 x$ odd?

Step-by-Step Reasoning

Recall the definitions

A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

A function is even if $f(-x) = f(x)$ .
Check the parity of each factor
- $\sin(-x) = -\sin x$ , so $\sin x$ is odd.
- $\cos(-x) = \cos x$ , so $\cos x$ is even.
Now raise them to powers:
- $(\sin x)^5$ : odd power of an odd function → still odd.
- $(\cos x)^4$ : even power of an even function → still even.
Combine the two

The product of an odd function and an even function is odd:

f(-x) = \sin^5(-x) \cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x).

Apply the symmetric interval property …

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Q.Evaluate $\int_{-1}^1 \sin^5 x \cos^4 x\, dx$

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 This Problem Is About Symmetry, Not Computation

Step-by-Step Reasoning

Method: Odd/Even Symmetry Over a Symmetric Interval

Steps

Common Mistakes

Q.Evaluate ∫−11sin⁡5xcos⁡4x dx\int_{-1}^1 \sin^5 x \cos^4 x\, dx∫−11​sin5xcos4xdx

Why This Problem Is About Symmetry, Not Computation

Step-by-Step Reasoning

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Q.Evaluate $\int_{-1}^1 \sin^5 x \cos^4 x\, dx$

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