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Question Number 30871 by ajfour last updated on 27/Feb/18

$$If2f\left(x\right)+f(-x)=\frac{1}{x}\sin (x-\frac{1}{x})$$$$Find{\int }_{1/e}^{e}f\left(x\right)dx.$$

Commented by abdo imad last updated on 27/Feb/18

$$letcompletetheworkofsir{mrw}_2$$$$withf\left(x\right)=\frac{1}{3x}sin(x-\frac{1}{x})wehave$$$${\int }_{\frac{1}{e}}^{e}f\left(x\right)dx=\frac{1}{3}{\int }_{\frac{1}{e}}^e\frac{1}{x}sin(x-\frac{1}{x})dxletusethech.x=\frac{1}{t}\Rightarrow$$$$3{\int }_{\frac{1}{e}}^{e}f\left(x\right)dx=-{\int }_{\frac{1}{e}}^{e}tsin(\frac{1}{t}-t)\frac{-dt}{t^2}={\int }_{\frac{1}{e}}^{e}tsin(\frac{1}{t}-t)dt$$$$=-{\int }_{\frac{1}{e}}^{e}tsin(t-\frac{1}{t})dt=-3{\int }_{\frac{1}{e}}^{e}f\left(x\right)dx\Rightarrow$$$$6{\int }_{\frac{1}{e}}^{e}f\left(x\right)dx\Rightarrow {\int }_{\frac{1}{e}}^{e}f\left(x\right)dx=0(ifweworkinacorps$$$$withcarecteristic\ne 6)$$

Commented by ajfour last updated on 28/Feb/18

$$ThankyouSir.$$

Answered by mrW2 last updated on 27/Feb/18

$$2f\left(x\right)+f(-x)=\frac{1}{x}\sin (x-\frac{1}{x})...\left(i\right)$$$$\Rightarrow 2f(-x)+f\left(x\right)=-\frac{1}{x}\sin (-x+\frac{1}{x})...\left(ii\right)$$$$2\left(i\right)-\left(ii\right):$$$$3f\left(x\right)=\frac{2}{x}\sin (x-\frac{1}{x})-\frac{1}{x}\sin (x-\frac{1}{x})$$$$\Rightarrow f\left(x\right)=\frac{1}{3x}\sin (x-\frac{1}{x})$$$${\int }_{1/e}^{e}f\left(x\right)dx=\frac{1}{3}\int \frac{1}{x}\sin (x-\frac{1}{x})dx$$$$letu=x-\frac{1}{x}$$$$u^2=x^2+\frac{1}{x^2}-2$$$$u^2+4=x^2+\frac{1}{x^2}+2={(x+\frac{1}{x})}^2$$$$\sqrt{u^2+2^2}=x+\frac{1}{x}$$$$du=(1+\frac{1}{x^2})dx=(x+\frac{1}{x})\frac{dx}{x}=\sqrt{u^2+2^2}\frac{dx}{x}$$$$\frac{dx}{x}=\frac{du}{\sqrt{u^2+2^2}}$$$${\int }_{1/e}^{e}f\left(x\right)dx=\frac{1}{3}\int \frac{1}{x}\sin (x-\frac{1}{x})dx$$$$=\frac{1}{3}{\int }_{\frac{1}{e}-e}^{e-\frac{1}{e}}\frac{\sin u}{\sqrt{u^2+2^2}}du$$$$=\frac{1}{3}{\int }_{\frac{1}{e}-e}^0\frac{\sin u}{\sqrt{u^2+2^2}}du+\frac{1}{3}{\int }_0^{e-\frac{1}{e}}\frac{\sin u}{\sqrt{u^2+2^2}}du$$$$=-\frac{1}{3}{\int }_0^{e-\frac{1}{e}}\frac{\sin u}{\sqrt{u^2+2^2}}du+\frac{1}{3}{\int }_0^{e-\frac{1}{e}}\frac{\sin u}{\sqrt{u^2+2^2}}du$$$$=0$$

Commented by abdo imad last updated on 27/Feb/18

$$siryouaremorenearfromvaluelookthat$$$$\frac{1}{e}-e=-(e-\frac{1}{e})andthefunctionu\to \frac{sinu}{\sqrt{u^2+4}}isoddso$$$$\int (...)du=0.$$

Commented by mrW2 last updated on 27/Feb/18

$$thankyousir!$$$$I{didn}^{\prime }tlookattherangeofvalues.$$$$Iwasconcentriatedinhowtosolve$$$$theintegral.Itishardindeed.$$

Commented by mrW2 last updated on 27/Feb/18

$$Canwesolveit,ifthequestionis$$$$find\int f\left(x\right)dx?$$

Commented by prof Abdo imad last updated on 28/Feb/18

$$perhapsiwilltry...$$

Commented by ajfour last updated on 28/Feb/18

$$ThankyouSir.$$

Commented by abdo imad last updated on 28/Feb/18

$$becauseyouarefamiliarwithdangerousroadslikeme....$$

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