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Question Number 46842 by maxmathsup by imad last updated on 01/Nov/18

$$find\int \frac{dx}{x\sqrt{x-x^2}}$$

Commented by maxmathsup by imad last updated on 01/Nov/18

$$wehavex\sqrt{x-x^2}=x\sqrt{-x^2+x}=x\sqrt{-(x^2-x)}=x\sqrt{-(x^2-2\frac{x}{2}+\frac{1}{4}-\frac{1}{4})}$$$$=x\sqrt{\frac{1}{4}-{(x-\frac{1}{2})}^2}sochangementx-\frac{1}{2}=\frac{1}{2}sintgive$$$$\int \frac{dx}{x\sqrt{x-x^2}}=\int \frac{1}{2(\frac{1}{2}+\frac{1}{2}sint)\frac{1}{2}cost}costdt$$$$=\int \frac{2dt}{1+sint}{=}_{tan\left(\frac{t}{2}\right)=u}\int \frac{2}{1+\frac{2u}{1+u^2}}\frac{2du}{1+u^2}$$$$=4\int \frac{du}{1+u^2+2u}=4\int \frac{du}{{(u+1)}^2}=-\frac{4}{u+1}+c=\frac{-4}{tan\left(\frac{t}{2}\right)+1}+cbut$$$$sin\left(t\right)=2x-1\Rightarrow t=arcsin(2x-1)\Rightarrow$$$$I=\frac{-4}{tan\left(\frac{arcsin(2x-1)}{2}\right)+1}+c.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

$$t=\frac{1}{x}dx=\frac{-1}{t^2}dt$$$$\int \frac{-dt}{t^2\times \frac{1}{t}\times \sqrt{\frac{1}{t}-\frac{1}{t^2}}}$$$$\int \frac{-dt}{t\times \sqrt{\frac{t-1}{t^2}}}$$$$\int \frac{-dt}{\sqrt{t-1}}$$$$=-1\times \frac{{(t-1)}^{\frac{1}{2}}}{\frac{1}{2}}+c$$$$=-2{(\frac{1}{x}-1)}^{\frac{1}{2}}+c$$

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