Question

The value of the definite integral $\int_{-\pi/2}^{\pi/2} (x^3 + x\cos x + \tan^5 x + 1) \, dx$ is:

• A. $\pi$
• B. $\frac{\pi}{2}$
• C. $0$
• D. $2\pi$

Hint:

Odd powers of $x$, $\tan x$, and combinations like $x \cos x$ are odd functions. They vanish completely over symmetric limits like $[-a, a]$.

Answer 1

The Correct Option is A

Step 1: Concept
We use the symmetric interval property for even and odd functions: \[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(x) \text{ is an odd function } (f(-x) = -f(x)) \]

Step 2: Meaning
We split the integrand into two parts: the sum of odd functions and a constant term.

Step 3: Analysis
Let the integral be split as follows: \[ I = \int_{-\pi/2}^{\pi/2} (x^3 + x\cos x + \tan^5 x) \, dx + \int_{-\pi/2}^{\pi/2} 1 \, dx \] Let $g(x) = x^3 + x\cos x + \tan^5 x$. Checking if $g(x)$ is odd: \[ g(-x) = (-x)^3 + (-x)\cos(-x) + \tan^5(-x) \] \[ g(-x) = -x^3 - x\cos x - \tan^5 x = -g(x) \] Since $g(x)$ is purely an odd function, its integral over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is zero: \[ \int_{-\pi/2}^{\pi/2} g(x) \, dx = 0 \] Now evaluate the second integral: \[ I = 0 + \int_{-\pi/2}^{\pi/2} 1 \, dx = [x]_{-\pi/2}^{\pi/2} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi \]

Step 4: Conclusion
The value of the definite integral is $\pi$. Final Answer: (A)