MCQ
$\int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} =$
- A$0$
- B$2\int_0^1 {\frac{{\sin x}}{{3 - |x|}}\,dx} $
- ✓$2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$
- D$2\int_0^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $
Here, $f(x) = \frac{{\sin x}}{{3 - |x|}}$ is an odd function but
$f(x) = \frac{{{x^2}}}{{3 - |x|}}$ is an even function
$\therefore \,\,I = - \int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}\,} dx = - 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$$ = 2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}\,} dx$.
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