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Question Number 171484 by infinityaction last updated on 16/Jun/22

$$$$$$letf\left(x\right)=x+\frac{2}{1.3}{x}^3+\frac{2.4}{1.3.5}{x}^5+\frac{2.4.6}{1.3.5.7}{x}^7+.........$$$$\mathrm{\forall }x\in (0,1)thevalueoff\left(\frac{1}{\sqrt{2}}\right)=?$$

Commented by mr W last updated on 16/Jun/22

$$igot$$$$f\left(x\right)=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$$$$f\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}\pi }{4}$$

Answered by mr W last updated on 16/Jun/22

$$f\left(x\right)=x+\frac{2}{1.3}{x}^3+\frac{2.4}{1.3.5}{x}^5+\frac{2.4.6}{1.3.5.7}{x}^7+...$$$$xf\left(x\right)=x^2+\frac{2}{1.3}{x}^4+\frac{2.4}{1.3.5}{x}^6+\frac{2.4.6}{1.3.5.7}{x}^8+...$$$${\int }_0^{x}xf\left(x\right)dx=\frac{1}{3}{x}^3+\frac{2}{1.3.5}{x}^5+\frac{2.4}{1.3.5.7}{x}^7+\frac{2.4.6}{1.3.5.7.9}{x}^9+...$$$$\frac{1}{x}{\int }_0^{x}xf\left(x\right)dx=\frac{1}{1.3}{x}^2+\frac{2}{1.3.5}{x}^4+\frac{2.4}{1.3.5.7}{x}^6+\frac{2.4.6}{1.3.5.7.9}{x}^8+...$$$${\left[\frac{1}{x}{\int }_0^{x}xf\left(x\right)dx\right]}^{\prime }=\frac{2}{1.3}x+\frac{2.4}{1.3.5}{x}^3+\frac{2.4.6}{1.3.5.7}{x}^5+\frac{2.4.6.8}{1.3.5.7.9}{x}^7+...$$$$x^2{\left[\frac{1}{x}{\int }_0^{x}xf\left(x\right)dx\right]}^{\prime }=\frac{2}{1.3}{x}^3+\frac{2.4}{1.3.5}{x}^5+\frac{2.4.6}{1.3.5.7}{x}^7+\frac{2.4.6.8}{1.3.5.7.9}{x}^9+...$$$$x^2{\left[\frac{1}{x}{\int }_0^{x}xf\left(x\right)dx\right]}^{\prime }=-x+x+\frac{2}{1.3}{x}^3+\frac{2.4}{1.3.5}{x}^5+\frac{2.4.6}{1.3.5.7}{x}^7+\frac{2.4.6.8}{1.3.5.7.9}{x}^9+...$$$$x^2{\left[\frac{1}{x}{\int }_0^{x}xf\left(x\right)dx\right]}^{\prime }=-x+f\left(x\right)$$$$x^2[-\frac{1}{x^2}{\int }_0^{x}xf\left(x\right)dx+f\left(x\right)]=-x+f\left(x\right)$$$${\int }_0^{x}xf\left(x\right)dx=(x^2-1)f\left(x\right)+x$$$$xf\left(x\right)=2xf\left(x\right)+(x^2-1)f^{\prime }\left(x\right)+1$$$$(x^2-1)f^{\prime }\left(x\right)+xf\left(x\right)+1=0$$$$\Rightarrow {\mathit{f}}^{\prime }\left(\mathit{x}\right)+\frac{\mathit{x}}{{\mathit{x}}^2-1}\mathit{f}\left(\mathit{x}\right)=\frac{1}{1-{\mathit{x}}^2}$$$$I=e^{\int \frac{xdx}{x^2-1}}=e^{\frac{\ln \mid 1-x^2\mid }{2}}=\sqrt{1-x^2}$$$$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{1-x^2}}\int \frac{\sqrt{1-x^2}}{1-x^2}dx=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}+C$$$$f\left(0\right)=0\Rightarrow C=0$$$$\Rightarrow f\left(x\right)=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$$$$\Rightarrow f\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}\pi }{4}$$

Commented by infinityaction last updated on 16/Jun/22

$$thankyousir$$$$nicesolution$$

Commented by Tawa11 last updated on 16/Jun/22

$$\mathrm{Great}\mathrm{sir}.$$

Answered by qaz last updated on 16/Jun/22

$$f\left(x\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{\left(2k\right)!!}{(2k+1)!!}{x}^{2k+1}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{x}^{2k+1}{\int }_0^{\pi /2}\mathrm{s}{in}^{2k+1}tdt={\int }_0^{\pi /2}\frac{x\sin t}{1-{\left(x\sin t\right)}^2}dt$$$$\Rightarrow f\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\frac{\sin t}{1-\frac{1}{2}\sin {}^{2}t}dt$$$$=\sqrt{2}{\int }_0^{\pi /2}\frac{\sin t}{1+\cos {}^{2}t}dt$$$$=\sqrt{2}\cdot \arctan \left(\cos t\right){\mid }_{\pi /2}^0$$$$=\frac{\sqrt{2}\pi }{4}$$

Commented by infinityaction last updated on 16/Jun/22

$$amazingsolutionsir$$

Answered by infinityaction last updated on 16/Jun/22

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