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Question-138485




Question Number 138485 by SLVR last updated on 14/Apr/21
Commented by SLVR last updated on 14/Apr/21
Good morning mr.W thanks for  your support .The above is also i  missed long ago..and i couldnot  retrive from the group.kindly help
$${Good}\:{morning}\:{mr}.{W}\:{thanks}\:{for} \\ $$$${your}\:{support}\:.{The}\:{above}\:{is}\:{also}\:{i} \\ $$$${missed}\:{long}\:{ago}..{and}\:{i}\:{couldnot} \\ $$$${retrive}\:{from}\:{the}\:{group}.{kindly}\:{help} \\ $$
Answered by phanphuoc last updated on 14/Apr/21
u=arcxsinx   dv=1/(√(1−x+x^2 ))dx
$${u}={arcxsinx}\:\:\:{dv}=\mathrm{1}/\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }{dx} \\ $$
Answered by Ñï= last updated on 14/Apr/21
∫_0 ^1 ((sin^(−1) (√x))/( (√(1−x+x^2 ))))dx=∫_0 ^1 ((sin^(−1) (√(1−x)))/( (√(1−x+x^2 ))))dx=(1/2)∙(π/2)∫_0 ^1 (dx/( (√((x−(1/2))^2 +(3/4)))))  =(π/4)ln(x−(1/2)+(√(1−x+x^2 )))_0 ^1 =(π/4)ln 3  {sin^(−1) (√x)+sin^(−1) (√(1−x))=(π/2),∫(du/( (√(u^2 +a^2 ))))=ln∣u+(√(u^2 +a^2 ))∣+C}  ∫_0 ^1 ((sin^(−1) x)/( (√(1−x+x^2 ))))dx=(π/4)ln 3
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{si}{n}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\right)_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}}{ln}\:\mathrm{3} \\ $$$$\left\{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}+\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}}=\frac{\pi}{\mathrm{2}},\int\frac{{du}}{\:\sqrt{{u}^{\mathrm{2}} +{a}^{\mathrm{2}} }}={ln}\mid{u}+\sqrt{{u}^{\mathrm{2}} +{a}^{\mathrm{2}} }\mid+{C}\right\} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}\cancel{=}\frac{\pi}{\mathrm{4}}{ln}\:\mathrm{3} \\ $$
Commented by SLVR last updated on 14/Apr/21
thanks sir... i am wrong writing  the question.
$${thanks}\:{sir}…\:{i}\:{am}\:{wrong}\:{writing} \\ $$$${the}\:{question}. \\ $$

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