MCQ
$\int\limits_{ - 2}^\pi  {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}} \,dx$ is equal to               (where $[·]$ denotes the greatest integer function)
  • A
    $\pi  + \sin 2\cos 2$
  • $\pi  - 2 + \sin 2\cos 2$
  • C
    $\pi  - 2 - \sin 2\cos 2$
  • D
    None

Answer

Correct option: B.
$\pi  - 2 + \sin 2\cos 2$
b
$\int_{-2}^{0} \frac{\sin ^{2} x}{-1+\frac{1}{2}} d x+\int_{0}^{\pi} \frac{\sin ^{2} x}{0+\frac{1}{2}} d x$

$-\int_{-2}^{0} 2 \sin ^{2} x d x+\int_{0}^{\pi} 2 \sin ^{2} x d x$

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