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Question Number 48500 by peter frank last updated on 24/Nov/18

Commented by Abdo msup. last updated on 24/Nov/18

$$y\left(x\right)=e^{arctan\left(x\right)}\Rightarrow y^{{}^{\prime }}\left(x\right)=\frac{1}{1+x^2}y\left(x\right)\Rightarrow$$$$y^{{}^{″}}\left(x\right)=-\frac{2x}{{(1+x^2)}^2}y\left(x\right)+\frac{1}{1+x^2}{y}^{{}^{\prime }}\left(x\right)$$$$=\frac{-2x}{{(1+x^2)}^2}y\left(x\right)+\frac{1}{{(1+x^2)}^2}y\left(x\right)=\frac{-2x+1}{{(1+x^2)}^2}y\left(x\right)\Rightarrow$$$$(1+x^2)y^{{}^{″}}-(2x-1)y^{{}^{\prime }}$$$$=\frac{(1+x^2)(-2x+1)}{{(1+x^2)}^2}y\left(x\right)-(2x-1)\frac{1}{1+x^2}y\left(x\right)$$$$=\frac{(1+x^2)(-2x+1)}{{(1+x^2)}^2}y\left(x\right)+\frac{(1-2x)(1+x^2)}{{(1+x^2)}^2}y\left(x\right)$$$$=0\Rightarrow (1+x^2)y^{{}^{″}}-(2x-1)y^{{}^{\prime }}=0sothereisaerrorat?the$$$$question!$$

Commented by Abdo msup. last updated on 24/Nov/18

$$letI={\int }_0^{\frac{1}{2}}arcosxdxbyparts$$$$I={\left[xarvosx\right]}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}\frac{-x}{\sqrt{1-x^2}}dx$$$$=\frac{\pi }{6}+{\int }_0^{\frac{1}{2}}\frac{xdx}{\sqrt{1-x^2}}chsngementx=sin\theta give$$$${\int }_0^{\frac{1}{2}}\frac{xdx}{\sqrt{1-x^2}}={\int }_0^{\frac{\pi }{6}}\frac{sin\theta }{cos\theta }cos\theta d\theta ={\int }_0^{\frac{\pi }{6}}sin\theta d\theta$$$$={[-cos\theta ]}_0^{\frac{\pi }{6}}=1-\frac{\sqrt{3}}{2}\Rightarrow I=\frac{\pi }{6}+1-\frac{\sqrt{3}}{2}$$

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