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Question Number 85909 by jagoll last updated on 26/Mar/20

$$\int {\sin }^{-1}\left(\sqrt{\frac{\mathrm{x}}{\mathrm{a}+\mathrm{x}}}\right)\mathrm{dx},\mathrm{a}>0$$

Commented byjohn santu last updated on 26/Mar/20

$$let\sqrt{\frac{x}{a+x}}=n\Rightarrow x=\frac{{an}^2}{1-n^2}$$ $$I=\int {\sin }^{-1}\left(n\right)\times \frac{2an}{{(1-n^2)}^2}dn$$ $$[h={\sin }^{-1}\left(n\right)\Rightarrow n=\sin h]$$ $$I=2a\int h\tan h\sec {}^{2}hdh$$ $$I=a(h\tan {}^{2}h-\int (\sec {}^{2}h-1)dh)$$ $$I=a(h\tan {}^{2}h-\tan h+h)+c$$ $$\therefore I=x{\sin }^{-1}\left(\sqrt{\frac{x}{a+x}}\right)-\sqrt{ax}+a{\sin }^{-1}\left(\sqrt{\frac{x}{a+x}}\right)+c$$

Commented byjagoll last updated on 26/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}$$

Answered by TANMAY PANACEA. last updated on 26/Mar/20

$$x={atan}^2\alpha \to \frac{dx}{d\alpha }=a\times 2tan\alpha \times {sec}^2\alpha$$ $$\sqrt{\frac{{atan}^2\alpha }{a+{atan}^2\alpha }}=\sqrt{\frac{{sin}^2\alpha }{{cos}^2\alpha \times {sec}^2\alpha }}=sin\alpha$$ $$I=\int \alpha .2atan\alpha .{sec}^2\alpha d\alpha$$ $$\frac{I}{2a}=\alpha \int tan\alpha {sec}^2\alpha d\alpha -\int [\frac{d\alpha }{d\alpha }\int tan\alpha {sec}^2\alpha d\alpha ]d\alpha$$ $$=\alpha .\frac{{tan}^2\alpha }{2}-\frac{1}{2}\int ({sec}^2\alpha -1)d\alpha$$ $$=\frac{\alpha {tan}^2\alpha }{2}-\frac{tan\alpha }{2}+\frac{\alpha }{2}$$ $$now\alpha ={tan}^{-1}\sqrt{\frac{x}{a}}$$ $$I=(\frac{{tan}^{-1}\sqrt{\frac{x}{a}}\times \frac{x}{a}}{2}-\frac{\sqrt{\frac{x}{a}}}{2}+\frac{{tan}^{-1}\sqrt{\frac{x}{a}}}{2})\times 2a$$ $$I={xtan}^{-1}\sqrt{\frac{x}{a}}-a\sqrt{\frac{x}{a}}+{atan}^{-1}\sqrt{\frac{x}{a}}$$ $$=(x+a){tan}^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}$$ $$plscheck$$

Answered by Kunal12588 last updated on 26/Mar/20

$$I={xsin}^{-1}\sqrt{\frac{x}{a+x}}-\int x\times \frac{1}{\sqrt{1-\frac{x}{a+x}}}\times \frac{1}{2\sqrt{\frac{x}{a+x}}}\times \frac{a+x-x}{{(a+x)}^2}dx$$ $$={xsin}^{-1}\sqrt{\frac{x}{a+x}}-\int x\times \frac{\sqrt{a+x}}{\sqrt{a}}\times \frac{\sqrt{a+x}}{2\sqrt{x}}\times \frac{a}{{(a+x)}^2}dx$$ $$={xsin}^{-1}\sqrt{\frac{x}{a+x}}-\frac{1}{2}\sqrt{a}\int \sqrt{x}\times \frac{1}{a+x}dx$$ $$I_1=\int \frac{\sqrt{x}}{a+x}dx$$ $$\sqrt{x}=t$$ $$dx=2\sqrt{x}dt$$ $$I_1=2\int \frac{x}{a+x}dt=2\int \frac{t^2}{a+t^2}dt$$ $$=2\int \frac{a+t^2}{a+t^2}dt-2a\int \frac{dt}{a+t^2}$$ $$=2t-2a\times \frac{1}{\sqrt{a}}{tan}^{-1}\frac{t}{\sqrt{a}}+c$$ $$=2\sqrt{x}-2\sqrt{a}{tan}^{-1}\frac{\sqrt{x}}{\sqrt{a}}+c$$ $$I={xsin}^{-1}\sqrt{\frac{x}{a+x}}-\frac{1}{2}\sqrt{a}(2\sqrt{x}-2\sqrt{a}{tan}^{-1}\sqrt{\frac{x}{a}})+C$$ $$I={xsin}^{-1}\sqrt{\frac{x}{a+x}}-\sqrt{ax}+a{tan}^{-1}\sqrt{\frac{x}{a}}+C$$

Commented byKunal12588 last updated on 26/Mar/20

$${sin}^{-1}\sqrt{\frac{x}{a+x}}={tan}^{-1}\sqrt{\frac{x}{a}}$$ $$soTanmaysiryouranswerisalsocorrect!$$

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