Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{2\pi}\cos^5 x\,dx$

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.

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

Why Does This Work? (A Quick Proof)

Split at zero:

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

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

Examples to Cement the Idea

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

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