Q.Choose the correct answer: The value of $\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5 x+1)\,dx$ is (A) 0 (B) 2 (C) $\pi$ (D) 1

The key idea is Definite Integral Symmetry: for an integral over $[ -a, a ]$, odd functions integrate to zero, while even functions contribute twice their integral over $[0, a]$.

Step 1: Split the integrand into odd and even parts.

$x^3$, $x\cos x$, and $\tan^5 x$ are all odd functions. The constant $1$ is even.

Step 2: The integral of each odd function over $[-\pi/2, \pi/2]$ is zero.

So $\int_{-\pi/2}^{\pi/2} (x^3 + x\cos x + \tan^5 x)\,dx = 0$. …

The key insight here is symmetry. When you integrate over $[-a, a]$, any odd function — a function $f(x)$ satisfying $f(-x) = -f(x)$ — contributes zero. That’s because the area on the left cancels the area on the right exactly. The given integrand is a sum of several terms, and most of them are odd. Only the constant term survives.

Let’s break it down.

1. Identify the odd terms.

   • $x^3$: $(-x)^3 = -x^3$, so it’s odd.
   • $x \cos x$: $\cos x$ is even, $x$ is odd, product is odd.
   • $\tan^5 x$: $\tan x$ is odd, so any odd power of it is odd. All three are odd functions.

2. The constant term.

   The $+1$ is even (in fact, it’s constant, so trivially even). Its integral over symmetric limits is just $1$ times the length of the interval.

3. Apply the odd-function property.

   For any odd function $f(x)$,

So:

$\int_{-\pi/2}^{\pi/2} x^3\,dx = 0,\quad \int_{-\pi/2}^{\pi/2} x\cos x\,dx = 0,\quad \int_{-\pi/2}^{\pi/2} \tan^5 x\,dx = 0.$ …

Method: Odd/even decomposition over symmetric limits

Over $[-a,a]$, split a sum into odd and even parts: odd terms vanish, and only the even terms contribute (twice their $[0,a]$ integral). This is the fastest route for a mixed polynomial-trig integrand.

Steps

Step 1: Classify each term's parity.

Odd powers of $x$, and products like $x\cos x$ or $\tan^{2k+1}x$, are odd; even powers and constants are even.

Step 2: Discard the odd terms.

$\int_{-a}^{a} (\text{odd})\,dx = 0.$ …

Common Mistakes

Mistake 1: Missing that $\tan^5x$ is odd.

Why it's wrong: $\tan(-x)=-\tan x$, so $\tan^5(-x)=-\tan^5x$ is odd and integrates to $0$ over $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$; keeping it wastes effort or invites error. Correct approach: classify it as odd and discard it.

Mistake 2: Forgetting the $+1$ contributes the whole answer. …