The value of integral _ 1/ ^ 2/ (1/x) x^2 ,dx = - Vidyadip

MCQ

The value of integral $\int_{1/\pi }^{2/\pi } {\frac{{\sin (1/x)}}{{{x^2}}}} \,dx = $

A
$2$
B
$ - 1$
C
$0$
✓
$1$

✓

Answer

Correct option: D.

$1$

d
(d) Put $t = \frac{1}{x} $

$\Rightarrow dt = - \frac{1}{{{x^2}}}dx$

as $t = \frac{\pi }{2}$ and $\pi $

$\therefore $ $\int_{1/\pi }^{2/\pi } {\frac{{\sin \left( {\frac{1}{x}} \right)}}{{{x^2}}}dx} $

$ = - \int_{\pi /2}^\pi {\sin t\,dt = - [\cos t]_{\pi /2}^\pi } $

$ = - \left[ {\cos \pi - \cos \left( {\frac{\pi }{2}} \right)} \right] = 1$.

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