Question

Evaluate \[ \int_{-1}^{1}\frac{\sin x-x^2}{3-|x|}\,dx = \]

• A. \(7+18\log \dfrac{3}{2}\)
• B. \(18\log \dfrac{9}{4}\)
• C. \(7+9\log \dfrac{9}{4}\)
• D. \(7-18\log \dfrac{3}{2}\)

Hint:

For definite integrals over symmetric limits, always check whether the integrand is odd or even before integrating.

Answer 1

The Correct Option is D

Step 1: Split the integral into odd and even parts.
Given, \[ I=\int_{-1}^{1}\frac{\sin x-x^2}{3-|x|}\,dx \] Now, \[ \sin x \] is an odd function and \[ 3-|x| \] is even. Hence, \[ \frac{\sin x}{3-|x|} \] is odd.
Therefore, \[ \int_{-1}^{1}\frac{\sin x}{3-|x|}\,dx=0 \] Thus, \[ I=-\int_{-1}^{1}\frac{x^2}{3-|x|}\,dx \] Since the integrand is even, \[ I=-2\int_{0}^{1}\frac{x^2}{3-x}\,dx \]

Step 2: Simplify the integrand.
Divide: \[ \frac{x^2}{3-x} = -x-3+\frac{9}{3-x} \] Hence, \[ I=-2\int_0^1\left(-x-3+\frac{9}{3-x}\right)\,dx \]

Step 3: Integrate term by term.
\[ I=-2\left[ -\frac{x^2}{2}-3x-9\log(3-x) \right]_0^1 \] Substitute limits: \[ I=-2\left[ -\frac{1}{2}-3-9\log2+9\log3 \right] \] \[ I=-2\left[ -\frac{7}{2}+9\log\frac{3}{2} \right] \] \[ I=7-18\log\frac{3}{2} \]

Step 4: Final conclusion.
Hence, \[ \boxed{ 7-18\log\frac{3}{2} } \]