Question Number 151587 by peter frank last updated on 22/Aug/21
$$\int\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\mathrm{x}}{\mathrm{a}+\mathrm{x}}}\:\mathrm{dx} \\ $$
Answered by MJS_new last updated on 22/Aug/21
$${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:\rightarrow\:{v}'=\frac{\sqrt{{a}}}{\mathrm{2}\left({x}+{a}\right)\sqrt{{x}}} \\ $$$$\int\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:{dx}= \\ $$$$={x}\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:−\frac{\sqrt{{a}}}{\mathrm{2}}\int\frac{\sqrt{{x}}}{{x}+{a}}{dx}= \\ $$$$={x}\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:−\sqrt{{ax}}+{a}\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:= \\ $$$$=−\sqrt{{ax}}+\left({x}+{a}\right)\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:+{C} \\ $$