Question Number 146901 by mathmax by abdo last updated on 16/Jul/21
$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{cos}\left(\mathrm{2arcsinx}\right)\:\: \\ $$$$\mathrm{calculate}\:\frac{\mathrm{dg}}{\mathrm{dx}}\:\mathrm{and}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{g}}{\mathrm{dx}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by ArielVyny last updated on 18/Jul/21
![(dg/dx)=−2(1/( (√(1−x^2 ))))sin(2arcsinx) (d^2 g/dx^2 )=−2[−(((−2x)/(2(√(1−x^2 ))))/(1−x^2 ))sin(2arcsinx)−2(1/( (√(1−x^2 ))))cos(2arsinx)] =−2[−((−x)/( (√(1−x^2 ))))×(1/(1−x^2 ))sin(2arsinx)−2(1/( (√(1−x^2 ))))cos(2arcsinx)] (d^2 g/dx^2 )=−2[((xsin(2arcsinx))/((1−x^2 )^(3/2) ))−2((cos(2arcsinx))/( (√(1−x^2 ))))] 2 ...∫_(−(1/2)) ^(1/2) cos[2arcsinx]dx f(x)=cos(2arcsinx) est paire donc ∫_(−(1/2)) ^(1/2) cos(2arcsinx)dx=2∫_0 ^(1/2) cos(2arcsinx)dx on pose du=1→u=x v=cos(2arcsinx)→dv=−2(1/( (√(1−x^2 ))))sin(2arcsinx) I=2[xcos(2arcsinx)]+2∫_0 ^(1/2) ((2x)/( (√(1−x^2 ))))sin(2arcsinx)dx du=((2x)/( (√(1−x^2 ))))→u=−(√(1−x^2 )) v=sin(2arcsinx)→dv=2(1/( (√(1−x^2 ))))cos(2arsinx) I=[2xcos(2arcsinx)]_0 ^1 −[2(√(1−x^2 ))sin(2arcsinx)]_0 ^1 +2∫_0 ^1 (2/( (√(1−x^2 ))))cos(2arcsinx) I=[2xcos(2arcsinx)−2(√(1−x^2 ))sin(2arcsinx)+2sin(2arcsinx)]_0 ^1 I=−2 Mr mathmax still chek my work please](https://www.tinkutara.com/question/Q147193.png)
$$\frac{{dg}}{{dx}}=−\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{sin}\left(\mathrm{2}{arcsinx}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {g}}{{dx}^{\mathrm{2}} }=−\mathrm{2}\left[−\frac{\frac{−\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}{\mathrm{1}−{x}^{\mathrm{2}} }{sin}\left(\mathrm{2}{arcsinx}\right)−\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{cos}\left(\mathrm{2}{arsinx}\right)\right] \\ $$$$=−\mathrm{2}\left[−\frac{−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }{sin}\left(\mathrm{2}{arsinx}\right)−\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{cos}\left(\mathrm{2}{arcsinx}\right)\right] \\ $$$$\frac{{d}^{\mathrm{2}} {g}}{{dx}^{\mathrm{2}} }=−\mathrm{2}\left[\frac{{xsin}\left(\mathrm{2}{arcsinx}\right)}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }−\mathrm{2}\frac{{cos}\left(\mathrm{2}{arcsinx}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right] \\ $$$$ \\ $$$$\mathrm{2}\:\:\:…\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {cos}\left[\mathrm{2}{arcsinx}\right]{dx} \\ $$$${f}\left({x}\right)={cos}\left(\mathrm{2}{arcsinx}\right)\:{est}\:{paire}\:{donc} \\ $$$$\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {cos}\left(\mathrm{2}{arcsinx}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {cos}\left(\mathrm{2}{arcsinx}\right){dx} \\ $$$${on}\:{pose}\:{du}=\mathrm{1}\rightarrow{u}={x} \\ $$$${v}={cos}\left(\mathrm{2}{arcsinx}\right)\rightarrow{dv}=−\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{sin}\left(\mathrm{2}{arcsinx}\right) \\ $$$${I}=\mathrm{2}\left[{xcos}\left(\mathrm{2}{arcsinx}\right)\right]+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{sin}\left(\mathrm{2}{arcsinx}\right){dx} \\ $$$${du}=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\rightarrow{u}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${v}={sin}\left(\mathrm{2}{arcsinx}\right)\rightarrow{dv}=\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{cos}\left(\mathrm{2}{arsinx}\right) \\ $$$${I}=\left[\mathrm{2}{xcos}\left(\mathrm{2}{arcsinx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{sin}\left(\mathrm{2}{arcsinx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{cos}\left(\mathrm{2}{arcsinx}\right) \\ $$$${I}=\left[\mathrm{2}{xcos}\left(\mathrm{2}{arcsinx}\right)−\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{sin}\left(\mathrm{2}{arcsinx}\right)+\mathrm{2}{sin}\left(\mathrm{2}{arcsinx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${I}=−\mathrm{2} \\ $$$${Mr}\:{mathmax}\:{still}\:{chek}\:{my}\:{work}\:\:{please} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$