MCQ
The value of integral $\int_{1/\pi }^{2/\pi } {\frac{{\sin (1/x)}}{{{x^2}}}} \,dx = $
- A$2$
- B$ - 1$
- C$0$
- ✓$1$
$\Rightarrow dt = - \frac{1}{{{x^2}}}dx$
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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