Integration Questions

Question Number 92668 by john santu last updated on 08/May/20

$$\underset{\mathrm{0}} {\overset{\mathrm{p}} {\int}}\:\sqrt{\frac{\mathrm{x}}{\mathrm{c}−\mathrm{x}}}\:\mathrm{dx}\:?\: \\$$

Commented byjohn santu last updated on 08/May/20

$$\mathrm{set}\:\mathrm{t}\:=\:\sqrt{\frac{{x}}{{c}−{x}}}\:,\:{x}\:=\:\frac{{ct}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:\: \\$$ $${dx}\:=\:\frac{\mathrm{2}{ct}\:{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\: \\$$ $${I}\:=\:\underset{\mathrm{0}} {\overset{\sqrt{\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}} {\int}}\frac{\:\:\mathrm{2}{ct}^{\mathrm{2}} \:{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\: \\$$ $${I}=\:−\mathrm{c}\:\left[\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}} +\:\mathrm{c}\left[\mathrm{tan}^{−\mathrm{1}} \:\mathrm{t}\right]_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}} \\$$ $$=−\mathrm{c}\:\left[\frac{\sqrt{\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}}{\mathrm{1}+\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}\:\right]\:+\mathrm{c}\:\left[\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{p}}{\mathrm{c}−\mathrm{p}}}\right]\: \\$$

Commented bymathmax by abdo last updated on 08/May/20

$${I}\:=\int_{\mathrm{0}} ^{{p}} \sqrt{\frac{{x}}{{c}−{x}}}{dx}\:\:{changement}\:{x}={csin}^{\mathrm{2}} {t}\:{give} \\$$ $${I}\:=\int_{\mathrm{0}} ^{{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)} \sqrt{\frac{{csin}^{\mathrm{2}} {t}}{{ccos}^{\mathrm{2}} {t}}}\mathrm{2}{csint}\:{cost} \\$$ $$=\mathrm{2}{c}\int_{\mathrm{0}} ^{{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)} \:\frac{{sint}}{{cost}}\:{sint}\:{cost}\:{dt}\:=\mathrm{2}{c}\:\int_{\mathrm{0}} ^{{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\$$ $$={c}\:{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)−\frac{{c}}{\mathrm{2}}\:\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)} \:+\lambda \\$$ $$={c}\:{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)−{c}\left[{sint}\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}\right]_{\mathrm{0}} ^{{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)} \:+\lambda \\$$ $$={c}\:{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)−{c}\sqrt{\frac{{p}}{{c}}}\sqrt{\mathrm{1}−\frac{{p}}{{c}}}\:+\lambda \\$$ $$={c}\:{arcsin}\left(\sqrt{\frac{{p}}{{c}}}\right)−\sqrt{{p}\left({c}−{p}\right)}\:+\lambda \\$$

Answered by behi83417@gmail.com last updated on 08/May/20

$$\mathrm{x}=\mathrm{c}.\mathrm{cos}^{\mathrm{2}} \mathrm{t}\Rightarrow\mathrm{dx}=−\mathrm{2csint}.\mathrm{costdt} \\$$ $$\Rightarrow\mathrm{I}=\int\frac{\mathrm{cost}}{\mathrm{sint}}.\left(−\mathrm{2c}.\mathrm{sint}.\mathrm{cost}\right)\mathrm{dt}= \\$$ $$=−\int\mathrm{2ccos}^{\mathrm{2}} \mathrm{tdt}=−\mathrm{c}\int\left(\mathrm{1}+\mathrm{cos2t}\right)\mathrm{dt}= \\$$ $$=−\mathrm{c}\left(\mathrm{t}+\mathrm{sint}.\mathrm{cost}\right)+\mathrm{const}.= \\$$ $$=−\mathrm{c}.\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{\mathrm{x}}{\mathrm{c}}\:}−\sqrt{\mathrm{cx}−\mathrm{x}^{\mathrm{2}} }+\mathrm{const}. \\$$ $$\left[\mathrm{I}=\underset{\mathrm{0}} {\overset{\mathrm{p}} {\int}}\sqrt{\frac{\mathrm{x}}{\mathrm{c}−\mathrm{x}}}=\frac{\boldsymbol{\mathrm{c}}.\pi}{\mathrm{2}}−\boldsymbol{\mathrm{c}}.\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \sqrt{\frac{\boldsymbol{\mathrm{p}}}{\boldsymbol{\mathrm{c}}}}−\sqrt{\boldsymbol{\mathrm{c}}.\boldsymbol{\mathrm{p}}−\boldsymbol{\mathrm{p}}^{\mathrm{2}} }\right] \\$$ $$\left[\mathrm{cost}=\sqrt{\frac{\mathrm{x}}{\mathrm{c}}},\mathrm{sint}=\sqrt{\mathrm{1}−\frac{\mathrm{x}}{\mathrm{c}}}\:\right] \\$$