Q65 / 125 · BITSAT 2020 · Mathematics

$\int_{\pi / 3}^{\pi / 2} x \sin (\pi[x]-x) \mathrm{dx}$ is equal to

  • A
    $\frac{1}{2}+\frac{\pi}{6}$
  • Correct answer
    B
    $1-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$
  • C
    $-\frac{1}{2}-\frac{\pi}{6}$
  • D
    $\frac{\sqrt{3}}{2}-1-\frac{\pi}{6}$

Solution

In the interval $\frac{\pi}{3}$ to $\frac{\pi}{2},[\mathrm{x}]=1$ $\begin{aligned} \therefore \mathrm{I} &=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin (\pi-\mathrm{x}) \mathrm{dx}=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin \mathrm{x} \mathrm{d} \mathrm{x} \\ &=[-\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}]_{\pi / 3}^{\pi / 2}=1-\frac{\sqrt{3}}{2}+\frac{\pi}{6} \end{aligned}$

Hence, the correct answer is option: 'B'.

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