Question Number 225994 by Spillover last updated on 17/Nov/25
Answered by Simurdiera last updated on 18/Nov/25
![Solucio^ n ∫_(1/2) ^( 1) ((√(1 − x^2 ))/(x∙sin(arccos(x)) + arcsin(x)))∙((√(1 − x^2 ))/( (√(1 − x^2 )))) dx ∫_(1/2) ^( 1) ((1 − x^2 )/((x∙sin(arccos(x)) + arcsin(x))(√(1 − x^2 )))) dx Cambio de variable u = arccos(x) ⇔ x = cos(u) du = − (1/( (√(1 − x^2 )))) dx Reemplazando −∫^ ((1 − cos^2 (u))/(cos(u)∙sin(u) + arcsin(cos(u)))) du −∫^ ((sin^2 (u))/(cos(u)∙sin(u) + arcsin(cos(u)))) du Cambio de variable m = cos(u)∙sin(u) + arcsin(cos(u)) dm = (− sin^2 (u) + cos^2 (u) − ((sin(u))/( (√(1 − cos^2 (u))))))du = (− sin^2 (u) + cos^2 (u) − 1)du = − 2∙sin^2 (u) du Reemplazando (1/2)∫(1/m) dm (1/2)∙ln(m) Regresando del cambio de variable (1/2)∙ln(cos(u)∙sin(u) + arcsin(cos(u))) Regresando del cambio de variable (1/2)∙ln(cos(arccos(x))∙sin(arccos(x)) + arcsin(cos(arccos(x)))) Aplicando algunas propiedades (1/2)∙ln(x∙sin(arccos(x)) + arcsin(x)) Evaluando los limites de integracio^ n (1/2)[ln(1∙sin(arccos(1)) + arcsin(1)) − ln((1/2)∙sin(arccos((1/2))) + arcsin((1/2)))] determinant ((((1/2)[ln((π/2)) − ln(((√3)/4) + (π/6))]))) Respuesta](https://www.tinkutara.com/question/Q226046.png)
$${Soluci}\acute {{o}n} \\ $$$$\int_{\mathrm{1}/\mathrm{2}} ^{\:\mathrm{1}} \frac{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}{{x}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left({x}\right)\right)\:+\:\mathrm{arcsin}\left({x}\right)}\centerdot\frac{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:{dx} \\ $$$$\int_{\mathrm{1}/\mathrm{2}} ^{\:\mathrm{1}} \frac{\mathrm{1}\:−\:{x}^{\mathrm{2}} }{\left({x}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left({x}\right)\right)\:+\:\mathrm{arcsin}\left({x}\right)\right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:{dx} \\ $$$$\mathrm{Cambio}\:\mathrm{de}\:\mathrm{variable}\: \\ $$$${u}\:=\:\mathrm{arccos}\left({x}\right)\:\:\:\:\Leftrightarrow\:\:\:\:\:{x}\:=\:\mathrm{cos}\left({u}\right) \\ $$$${du}\:=\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:{dx} \\ $$$$\mathrm{Reemplazando} \\ $$$$−\int^{\:} \frac{\mathrm{1}\:−\:\mathrm{cos}^{\mathrm{2}} \left({u}\right)}{\mathrm{cos}\left({u}\right)\centerdot\mathrm{sin}\left({u}\right)\:+\:\mathrm{arcsin}\left(\mathrm{cos}\left({u}\right)\right)}\:{du} \\ $$$$−\int^{\:} \frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{\mathrm{cos}\left({u}\right)\centerdot\mathrm{sin}\left({u}\right)\:+\:\mathrm{arcsin}\left(\mathrm{cos}\left({u}\right)\right)}\:{du} \\ $$$$\mathrm{Cambio}\:\mathrm{de}\:\mathrm{variable} \\ $$$${m}\:=\:\mathrm{cos}\left({u}\right)\centerdot\mathrm{sin}\left({u}\right)\:+\:\mathrm{arcsin}\left(\mathrm{cos}\left({u}\right)\right) \\ $$$${dm}\:=\:\left(−\:\mathrm{sin}^{\mathrm{2}} \left({u}\right)\:+\:\mathrm{cos}^{\mathrm{2}} \left({u}\right)\:−\:\frac{\mathrm{sin}\left({u}\right)}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}^{\mathrm{2}} \left({u}\right)}}\right){du}\:=\:\left(−\:\mathrm{sin}^{\mathrm{2}} \left({u}\right)\:+\:\mathrm{cos}^{\mathrm{2}} \left({u}\right)\:−\:\mathrm{1}\right){du}\:=\:−\:\mathrm{2}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)\:{du} \\ $$$$\mathrm{Reemplazando} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{m}}\:{dm} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{ln}\left({m}\right) \\ $$$$\mathrm{Regresando}\:\mathrm{del}\:\mathrm{cambio}\:\mathrm{de}\:\mathrm{variable}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{ln}\left(\mathrm{cos}\left({u}\right)\centerdot\mathrm{sin}\left({u}\right)\:+\:\mathrm{arcsin}\left(\mathrm{cos}\left({u}\right)\right)\right) \\ $$$$\mathrm{Regresando}\:\mathrm{del}\:\mathrm{cambio}\:\mathrm{de}\:\mathrm{variable} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{arccos}\left({x}\right)\right)\centerdot\mathrm{sin}\left(\mathrm{arccos}\left({x}\right)\right)\:+\:\mathrm{arcsin}\left(\mathrm{cos}\left(\mathrm{arccos}\left({x}\right)\right)\right)\right) \\ $$$$\mathrm{Aplicando}\:\mathrm{algunas}\:\mathrm{propiedades} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{ln}\left({x}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left({x}\right)\right)\:+\:\mathrm{arcsin}\left({x}\right)\right) \\ $$$$\mathrm{Evaluando}\:\mathrm{los}\:\mathrm{limites}\:\mathrm{de}\:\mathrm{integraci}\acute {\mathrm{o}n} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{1}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left(\mathrm{1}\right)\right)\:+\:\mathrm{arcsin}\left(\mathrm{1}\right)\right)\:−\:\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:+\:\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\right] \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)\:−\:\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:+\:\frac{\pi}{\mathrm{6}}\right)\right]}\\\hline\end{array}\:\:\:\boldsymbol{{Respuesta}} \\ $$
Answered by Spillover last updated on 19/Nov/25
