Question Number 5878 by gourav~ last updated on 03/Jun/16
$$\int\sqrt{\frac{{a}−{x}}{{a}+{x}}}{dx} \\ $$$${please}\:{help}…….= \\ $$
Answered by Yozzii last updated on 03/Jun/16
$${Let}\:{I}=\int\sqrt{\frac{{a}−{x}}{{a}+{x}}}{dx}. \\ $$$$\frac{{a}−{x}}{{a}+{x}}=\frac{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{ax}}={j} \\ $$$${Let}\:{x}={acosu}\Rightarrow\frac{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{cos}^{\mathrm{2}} {u}\right)}{{a}^{\mathrm{2}} \left(\mathrm{1}+{cosu}\right)^{\mathrm{2}} }\:\:\left({a}>\mathrm{0}\right) \\ $$$${j}=\frac{{sin}^{\mathrm{2}} {u}}{\left(\mathrm{1}+{cosu}\right)^{\mathrm{2}} }\Rightarrow\sqrt{{j}}=\sqrt{\frac{{a}−{x}}{{a}+{x}}}=\sqrt{\frac{{sin}^{\mathrm{2}} {u}}{\left(\mathrm{1}+{cosu}\right)^{\mathrm{2}} }} \\ $$$${If}\:{I}\in\mathbb{R}\Rightarrow\frac{{a}−{x}}{{a}+{x}}\geqslant\mathrm{0}\Rightarrow\frac{{x}−{a}}{{x}+{a}}\leqslant\mathrm{0}\Rightarrow−{a}\leqslant{x}\leqslant{a}\Rightarrow−\mathrm{1}\leqslant{cosu}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant{cosu}+\mathrm{1}\leqslant\mathrm{2}\:\therefore\:\sqrt{\left(\mathrm{1}+{cosu}\right)^{\mathrm{2}} }=\mathrm{1}+{cosu} \\ $$$$\therefore\:\mathrm{0}\leqslant{cos}^{\mathrm{2}} {u}\leqslant\mathrm{1} \\ $$$$−\mathrm{1}\leqslant−{cos}^{\mathrm{2}} {u}\leqslant\mathrm{0} \\ $$$$\mathrm{0}\leqslant\mathrm{1}−{cos}^{\mathrm{2}} {u}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {u}}\leqslant\mathrm{1}\Rightarrow\mathrm{0}\leqslant{sinu}\leqslant\mathrm{1} \\ $$$$\therefore\sqrt{{sin}^{\mathrm{2}} {u}}={sinu}. \\ $$$${In}\:{all},\:\sqrt{{j}}=\frac{{sinu}}{\mathrm{1}+{cosu}}. \\ $$$${x}={acosu}\Rightarrow{dx}=−{asinudu}. \\ $$$$\therefore{I}=\int\frac{−{asin}^{\mathrm{2}} {u}}{\mathrm{1}+{cosu}}{du}.\:\:\: \\ $$$${I}=\frac{−{a}}{\mathrm{2}}\int\frac{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{ucos}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{u}}{{cos}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{u}}{du}\:\:\:\left({alternatively},\frac{{sin}^{\mathrm{2}} {u}}{\mathrm{1}+{cosu}}=\frac{{sin}^{\mathrm{2}} {u}\left(\mathrm{1}−{cosu}\right)}{\left(\mathrm{1}+{cosu}\right)\left(\mathrm{1}−{cosu}\right)}=\frac{{sin}^{\mathrm{2}} {u}\left(\mathrm{1}−{cosu}\right)}{\mathrm{1}−{cos}^{\mathrm{2}} {u}}=\frac{{sin}^{\mathrm{2}} {u}\left(\mathrm{1}−{cosu}\right)}{{sin}^{\mathrm{2}} {u}}=\mathrm{1}−{cosu}\right) \\ $$$${I}=−\mathrm{2}{a}\int{sin}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{udu} \\ $$$${I}=−\mathrm{2}{a}\int\mathrm{0}.\mathrm{5}\left(\mathrm{1}−{cosu}\right){du} \\ $$$${I}=−{a}\int\left(\mathrm{1}−{cosu}\right){du} \\ $$$${I}=−{a}\left({u}−{sinu}\right)+{C} \\ $$$${I}={a}\left(\frac{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{a}}−{cos}^{−\mathrm{1}} \frac{{x}}{{a}}\right)+{C} \\ $$$${I}=\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{acos}^{−\mathrm{1}} \frac{{x}}{{a}}+{C} \\ $$
Commented by Yozzii last updated on 03/Jun/16
$${Alternatively}, \\ $$$${Let}\:\sqrt{{x}+\mathrm{2}}=\mathrm{2}{sinu} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}{dx}=\mathrm{2}{cosudu} \\ $$$${dx}=\mathrm{8}{sinucosudu} \\ $$$${x}+\mathrm{2}=\mathrm{4}{sin}^{\mathrm{2}} {u} \\ $$$${x}=\mathrm{4}{sin}^{\mathrm{2}} {u}−\mathrm{2} \\ $$$$−{x}=\mathrm{2}−\mathrm{4}{sin}^{\mathrm{2}} {u} \\ $$$$\mathrm{2}−{x}=\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} {u}=\mathrm{4}{cos}^{\mathrm{2}} {u}\Rightarrow{cosu}=\frac{\sqrt{\mathrm{2}−{x}}}{\mathrm{2}} \\ $$$${I}=\int\frac{\mathrm{2}{cosu}}{\mathrm{2}{sinu}}×\mathrm{8}{sinucosudu} \\ $$$${I}=\mathrm{8}\int{cos}^{\mathrm{2}} {udu} \\ $$$${I}=\mathrm{4}\int\left(\mathrm{1}+{cos}\mathrm{2}{u}\right){du} \\ $$$${I}=\mathrm{4}\left({u}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{u}\right)+{C} \\ $$$${I}=\mathrm{4}{u}+\mathrm{4}{sinucosu}+{C} \\ $$$${I}=\mathrm{4}{sin}^{−\mathrm{1}} \frac{\sqrt{{x}+\mathrm{2}}}{\mathrm{2}}+\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+{C} \\ $$