# I-0-1-c-x-2-x-1-x-2-dx-c-gt-1-

Question Number 68481 by ajfour last updated on 11/Sep/19
$${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{{c}−{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}{dx}\:\:\:\:\:\:\left({c}\:>\mathrm{1}\right) \\$$
Commented by MJS last updated on 11/Sep/19
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{everything}\:\mathrm{I}\:\mathrm{know},\:\mathrm{it}\:\mathrm{seems}\:\mathrm{impossible} \\$$$$\mathrm{to}\:\mathrm{solve} \\$$
Commented by Kunal12588 last updated on 12/Sep/19
Commented by Kunal12588 last updated on 12/Sep/19
Commented by MJS last updated on 12/Sep/19
$$\frac{\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\sqrt{\frac{{c}−{x}^{\mathrm{2}} }{{x}−{x}^{\mathrm{3}} }}}{\:\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{c}}}}=\mathrm{2}\sqrt{{cx}} \\$$$$\Rightarrow\:\left[\mathrm{2}\sqrt{{cx}}\mathrm{F}_{\mathrm{1}} \left(…\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}\sqrt{{c}}\mathrm{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}};\:−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{5}}{\mathrm{4}};\:\frac{\mathrm{1}}{{c}},\:\mathrm{1}\right) \\$$
Commented by MJS last updated on 12/Sep/19
$$\mathrm{Sir}\:\mathrm{Kunal},\:\mathrm{can}\:\mathrm{you}\:\mathrm{calculate}\:\mathrm{it}\:\mathrm{now}? \\$$
Commented by MJS last updated on 12/Sep/19
$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}: \\$$$$\mathrm{F}_{\mathrm{1}} \left({a},\:{b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} ,\:{c};\:{x};\:{y}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({a}\right)\Gamma\left({c}−{a}\right)}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{c}−{a}−\mathrm{1}} \left(\mathrm{1}−{xt}\right)^{−{b}_{\mathrm{1}} } \left(\mathrm{1}−{yt}\right)^{−{b}_{\mathrm{2}} } {dt} \\$$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\$$$$\frac{\sqrt{{c}}}{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\sqrt{{c}−{t}}}{{t}^{\mathrm{3}/\mathrm{4}} \sqrt{{c}\left(\mathrm{1}−{t}\right)}}{dt} \\$$$$…\mathrm{but}\:\mathrm{is}\:\mathrm{this}\:\mathrm{easier}? \\$$