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sin-1-x-a-x-dx-




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

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