Q.By using the properties of definite integrals, evaluate the integral $\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx$

The key idea is the Even Function Property: for an odd function $f$, $\int_{-a}^{a} f(x)\,dx = 0$.

1. Check if $\sin^7 x$ is odd. Since $\sin(-x) = -\sin x$, we have $\sin^7(-x) = (-\sin x)^7 = -\sin^7 x$. So $\sin^7 x$ is an odd function.

2. The limits are symmetric about zero: from $-\pi/2$ to $\pi/2$. …

Why this works — the Even Function Property

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 mechanical trick — it's about symmetry. An odd function satisfies $f(-x) = -f(x)$ for all $x$ in its domain. Graphically, it means the function is symmetric about the origin: whatever it does on the right side, it does the exact opposite on the left.

For a definite integral over symmetric limits $[-a, a]$, an odd function's area on the left is the negative of its area on the right. They cancel perfectly, giving zero. This is one of the most powerful shortcuts in integral calculus — it saves you from doing any actual integration.

Now, is $\sin^7 x$ odd? Let's check.

1. Check the parity of $\sin^7 x$

   Recall that $\sin x$ itself is odd: $\sin(-x) = -\sin x$.

   Raising an odd function to any odd power preserves oddness:

   $\sin^7(-x) = [\sin(-x)]^7 = (-\sin x)^7 = -\sin^7 x$.

   So yes, $\sin^7 x$ is odd.

2. Apply the symmetric-limit property

   Since the limits are $-\pi/2$ to $\pi/2$, which are symmetric about zero, and the integrand is odd, the integral must be zero.

   No computation of antiderivatives, no trigonometric identities, no substitution — just the symmetry argument.

3. Confirm with a quick mental check …

Method: Odd-function symmetry over symmetric limits

An odd integrand over $[-a,a]$ integrates to zero because the left and right areas cancel — recognising this avoids all computation.

Steps

Step 1: Test for oddness.

Check $f(-x)=-f(x)$. An odd power of $\sin$ is odd, since $\sin(-x)=-\sin x$ and an odd exponent keeps the minus sign.

Step 2: Confirm the limits are symmetric about $0$. …

Common Mistakes

Mistake 1: Attempting a reduction formula for $\sin^7x$.

Why it's wrong: over the symmetric interval $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ the odd symmetry gives $0$ instantly, so all that work is wasted (and invites errors). Correct approach: note $\sin^7(-x)=-\sin^7x$ and conclude the integral is $0$.

Mistake 2: Assuming any $\sin$ power vanishes over symmetric limits. …