Q.Prove that $\int_{-1}^{1}x^{17}\cos^4 x\,dx=0$

The key idea is odd-function symmetry under a definite integral over $[-a, a]$.

1. Let $f(x) = x^{17} \cos^4 x$. Check parity: $f(-x) = (-x)^{17} \cos^4(-x) = -x^{17} \cos^4 x = -f(x)$. Since $\cos x$ is even, $\cos^4 x$ is even; $x^{17}$ is odd, so their product is odd. …

The key to this problem is recognising symmetry — specifically, the property of definite integrals over intervals symmetric about zero. When you see an integral from $-a$ to $a$, your first instinct should be to check whether the integrand is odd or even. This isn't just a trick; it's a fundamental shortcut that saves you from grinding through a messy computation.

Let’s break it down.

1. Recall the definitions.

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

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

   The classic property:

   For an odd function $f$, $\int_{-a}^{a} f(x) \, dx = 0$.

   For an even function $f$, $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.

2. Examine each factor in the integrand.

   The integrand is $x^{17} \cos^4 x$.

   • $x^{17}$: Since $(-x)^{17} = -x^{17}$, this is an odd function.
   • $\cos^4 x$: Cosine is even ($\cos(-x) = \cos x$), so $\cos^4(-x) = (\cos(-x))^4 = (\cos x)^4 = \cos^4 x$. This is an even function.

3. What happens when you multiply an odd function by an even function?

   Let $f(x) = x^{17}$ (odd) and $g(x) = \cos^4 x$ (even). Their product $h(x) = f(x) \cdot g(x)$ satisfies:

So $h(x)$ is odd. …

Method: Odd/even symmetry over a symmetric interval

Use this for $\int_{-a}^{a} h(x)\,dx$: determine the parity of $h$ first — an odd integrand makes the integral $0$ with no computation.

Steps

Step 1: Test each factor's parity.

$x^{n}$ is odd for odd $n$, even for even $n$; $\cos^k x$ is even (since $\cos$ is even). Use the rules: odd$\times$even $=$ odd, odd$\times$odd $=$ even, even$\times$even $=$ even.

Step 2: Apply the symmetry result. …

Common Mistakes

Mistake 1: Thinking $\cos^4x$ being even cancels the oddness of $x^{17}$.

Why it's wrong: even $\times$ odd $=$ odd; an even factor preserves sign, it does not neutralise an odd factor. Correct approach: the product $x^{17}\cos^4x$ is odd, so the integral is $0$.

Mistake 2: Getting the parity product rule wrong. …