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Question Number 17008 by arnabpapu550@gmail.com last updated on 29/Jun/17

∫_((1 )/Π) ^(2/Π)  (1/x^2 )sin(1/x)dx

$$\int_{\frac{\mathrm{1}\:}{\Pi}} ^{\frac{\mathrm{2}}{\Pi}} \:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{sin}\frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx} \\ $$

Answered by sma3l2996 last updated on 29/Jun/17

t=(1/x)⇒dt=((−dx)/x^2 )  ∫_(1/π) ^(2/π) (1/x^2 )sin((1/x))dx=−∫_π ^(π/2) sin(t)dt=[cos(t)]_π ^(π/2) =1

$${t}=\frac{\mathrm{1}}{{x}}\Rightarrow{dt}=\frac{−{dx}}{{x}^{\mathrm{2}} } \\ $$$$\int_{\frac{\mathrm{1}}{\pi}} ^{\frac{\mathrm{2}}{\pi}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }{sin}\left(\frac{\mathrm{1}}{{x}}\right){dx}=−\int_{\pi} ^{\frac{\pi}{\mathrm{2}}} {sin}\left({t}\right){dt}=\left[{cos}\left({t}\right)\right]_{\pi} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1} \\ $$

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