MCQ
The value of integral $\int_{1 / \pi}^{2 / \pi} \frac{\sin \left(\frac{1}{x}\right)}{x^2} d x=$
  • A
    2
  • B
    -1
  • C
    $0$
  • 1

Answer

Correct option: D.
1
(D)
Put $\frac{1}{x}= t \Rightarrow-\frac{1}{x^2} d x= dt$
When $x=\frac{1}{\pi}, t =\pi$ and when $x=\frac{2}{\pi}, t =\frac{\pi}{2}$
$\begin{aligned} \therefore \quad \int_{1 / \pi}^{2 / \pi} \frac{\sin \left(\frac{1}{x}\right)}{x^2} d x & =-\int_\pi^{\pi / 2} \sin dt =[\cos t ]_\pi^{\pi / 2} \\ & =0-(-1)=1\end{aligned}$

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