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Question Number 92397 by john santu last updated on 06/May/20

$$\int \frac{1}{x-\sqrt{1-x^2}}dx$$$$[x=\sin w]$$$$\int \frac{\cos \mathrm{w}\mathrm{dw}}{\sin \mathrm{w}-\cos \mathrm{w}}=\int \frac{\mathrm{dw}}{\tan \mathrm{w}-1}$$$$=\int \frac{{\sec }^2\mathrm{w}\mathrm{dw}}{(\tan \mathrm{w}-1){\sec }^2\mathrm{w}}$$$$=\int \frac{\mathrm{du}}{(\mathrm{u}-1)({\mathrm{u}}^2+1)};[\mathrm{u}=\tan \mathrm{w}]$$$$=\int \frac{\mathrm{du}}{2(\mathrm{u}-1)}-\int \frac{\mathrm{u}\mathrm{du}}{2({\mathrm{u}}^2+1)}$$$$=\frac{1}{2}\ln \mid \mathrm{u}-1\mid -\frac{1}{4}\ln \mid {\mathrm{u}}^2+1\mid -\frac{1}{2}{\tan }^{-1}\left(\mathrm{u}\right)+\mathrm{c}$$$$=\frac{1}{2}\ln \mid \tan \mathrm{w}-1\mid -\frac{1}{4}\ln \mid \tan {}^2\mathrm{w}+1\mid -$$$$\frac{1}{2}{\tan }^{-1}\left(\tan \mathrm{w}\right)+\mathrm{c}$$$$=\frac{1}{2}\ln \mid \frac{x}{\sqrt{1-x^2}}-1\mid +\frac{1}{4}\ln \mid 1-x^2\mid -$$$$\frac{1}{2}{\sin }^{-1}\left(x\right)+c$$

Commented by abdomathmax last updated on 07/May/20

$$I=\int \frac{dx}{x-\sqrt{1-x^2}}wedothechangementx=sint\Rightarrow$$$$I=\int \frac{cost}{sint-cost}dt{=}_{tan\left(\frac{t}{2}\right)=u}\int \frac{\frac{1-u^2}{1+u^2}}{\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}\times \frac{2du}{1+u^2}$$$$=\int \frac{1-u^2}{(1+u^2)(2u-1+u^2)}du$$$$=-\int \frac{u^2-1}{(u^2+1)(u^2+2u-1)}duletdecompose$$$$F\left(u\right)=\frac{u^2-1}{(u^2+1)(u^2+2u-1)}$$$$u^2+2u-1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=2\Rightarrow u_1=-1+\sqrt{2}and$$$$u_2=-1-\sqrt{2}\Rightarrow F\left(u\right)=\frac{u^2-1}{(u-u_1)(u-u_2)(u^2+1)}$$$$=\frac{a}{u-u_1}+\frac{b}{u-u_2}+\frac{cu+d}{u^2+1}$$$$a=\frac{u_1^2-1}{2\sqrt{2}(u_1^2+1)},b=-\frac{u_2^2-1}{2\sqrt{2}(u_2^2+1)}$$$${lim}_{u\to 0}uF\left(u\right)=0=a+b+c\Rightarrow c=-a-b$$$$F\left(u\right)=\frac{a}{u-u_1}+\frac{b}{u-u_2}+\frac{(-a-b)u+d}{u^2+1}$$$$F\left(0\right)=1=-\frac{a}{u_1}-\frac{b}{u_2}+d\Rightarrow d=1+\frac{a}{u_1}+\frac{b}{u_2}$$$$\int F\left(u\right)du=aln\mid u-u_1\mid +bln\mid u-u_2\mid +\frac{c}{2}ln(u^2+1)$$$$+darctanu+\lambda$$$$=aln\mid tan\left(\frac{1}{2}arcsinx\right)-u_1\mid +bln\mid tan\left(\frac{1}{2}arcsinx\right)\mid$$$$+\frac{c}{2}ln(1+{tan}^2\left(\frac{1}{2}arcsinx\right))+darctan\left(tan\left(\frac{1}{2}arcsinx\right)\right)+\lambda$$$$resttosimplifya,b,c,d...$$$$$$$$$$$$$$

Commented by john santu last updated on 07/May/20

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