Question

Evaluate \[ \int_{-1}^{1}\frac{x^{2}\sin x}{x^{4}+1}\,dx. \]

• A. \(\dfrac{\pi}{2}\)
• B. \(\tan^{-1}2\)
• C. \(0\)
• D. \(1\)

Hint:

For definite integrals over symmetric limits: \[ \int_{-a}^{a}\text{Odd Function}\,dx=0, \] \[ \int_{-a}^{a}\text{Even Function}\,dx = 2\int_{0}^{a}\text{Function}\,dx. \] Always check parity before attempting integration.

Answer 1

The Correct Option is C

Concept: A very useful property of definite integrals is: \[ \int_{-a}^{a}f(x)\,dx=0 \] whenever \(f(x)\) is an odd function. Therefore the first step is to determine whether the integrand is odd or even.

Step 1: Define the integrand.
Let \[ f(x)=\frac{x^{2}\sin x}{x^{4}+1}. \] We examine the parity of this function.

Step 2: Compute \(f(-x)\).
Substituting \(-x\), \[ f(-x) = \frac{(-x)^2\sin(-x)} {(-x)^4+1}. \] Using \[ (-x)^2=x^2, \] \[ (-x)^4=x^4, \] and \[ \sin(-x)=-\sin x, \] we obtain \[ f(-x) = \frac{x^2(-\sin x)} {x^4+1}. \] \[ f(-x) = -\frac{x^2\sin x}{x^4+1}. \] \[ f(-x)=-f(x). \]

Step 3: Identify the nature of the function.
Since \[ f(-x)=-f(x), \] the integrand is an odd function.

Step 4: Apply the odd function property.
For any odd function, \[ \int_{-a}^{a}f(x)\,dx=0. \] Hence, \[ \int_{-1}^{1} \frac{x^{2}\sin x} {x^{4}+1} \,dx = 0. \] \[ \boxed{0} \]

Step 5: Select the correct option.
Therefore the required value is \[ \boxed{0} \] and option (C) is correct.