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Question Number 137508 by EDWIN88 last updated on 03/Apr/21

$$\int \frac{\sin \left(\frac{1}{x}\right)}{x^3}dx=?$$

Answered by liberty last updated on 03/Apr/21

$$L=\int \frac{1}{x}\sin \left(\frac{1}{x}\right)\left(\frac{1}{x^2}\right)dx$$$$lett=\frac{1}{x}\Rightarrow -dt=\frac{dx}{x^2}$$$$L=\int -t\sin tdt$$$$byparts\{\begin{array}{l}u=-t,du=-dt\\ v=-\cos t\end{array}$$$$L=t\cos t-\int \cos tdt$$$$L=t\cos t-\sin t+c$$$$L=\frac{\cos \left(\frac{1}{x}\right)}{x}-\sin \left(\frac{1}{x}\right)+c$$

Answered by mathmax by abdo last updated on 03/Apr/21

$$\mathrm{I}=\int \frac{\sin \left(\frac{1}{\mathrm{x}}\right)}{{\mathrm{x}}^3}\mathrm{dx}\Rightarrow \mathrm{I}{=}_{\frac{1}{\mathrm{x}}=\mathrm{t}}-\int \frac{\sin \left(\mathrm{t}\right)}{{\mathrm{t}}^2}.{\mathrm{t}}^3\mathrm{dt}=-\int \mathrm{tsint}\mathrm{dt}$$$$\mathrm{and}\int \mathrm{tsint}\mathrm{dt}{=}_{\mathrm{by}\mathrm{parts}}-\mathrm{tcost}+\int \mathrm{cost}\mathrm{dt}$$$$=-\mathrm{tcost}+\mathrm{sint}+\mathrm{C}=-\frac{1}{\mathrm{x}}\cos \left(\frac{1}{\mathrm{x}}\right)+\sin \left(\frac{1}{\mathrm{x}}\right)+\mathrm{C}$$

Commented by mathmax by abdo last updated on 03/Apr/21

$$\Rightarrow \mathrm{I}=\frac{1}{\mathrm{x}}\cos \left(\frac{1}{\mathrm{x}}\right)-\sin \left(\frac{1}{\mathrm{x}}\right)+\mathrm{C}$$

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