Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 39441 by rahul 19 last updated on 06/Jul/18

$${\int }_{\frac{1}{4}}^4\frac{1}{x}\sin (x-\frac{1}{x})dx=?$$

Commented by prof Abdo imad last updated on 06/Jul/18

$$changementx=\frac{1}{t}give$$$$I=-{\int }_{\frac{1}{4}}^{4}tsin(\frac{1}{t}-t)\frac{-dt}{t^2}$$$$={\int }_{\frac{1}{4}}^4\frac{1}{t}sin(\frac{1}{t}-t)dt=-{\int }_{\frac{1}{4}}^4\frac{1}{t}sin(t-\frac{1}{t})dt$$$$=-I\Rightarrow 2I=0\Rightarrow I=0$$

Commented by rahul 19 last updated on 06/Jul/18

Thanks prof Abdo ����

Terms of Service

Privacy Policy

Contact: info@tinkutara.com