Question Number 97537 by M±th+et+s last updated on 08/Jun/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\left({yx}^{\mathrm{3}} +{x}^{\mathrm{2}} −{yx}−\mathrm{1}\right)}{dx} \\ $$
Answered by smridha last updated on 08/Jun/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\boldsymbol{{yx}}+\mathrm{1}\right)\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}\:\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{sinA}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\boldsymbol{{dA}}}{\boldsymbol{{ysinA}}+\mathrm{1}}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\boldsymbol{{d}}\left(\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+\boldsymbol{{y}}\right)}{\left(\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+{y}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\boldsymbol{{y}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{y}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\right)\right. \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\left(\mathrm{1}+\boldsymbol{{y}}\right)\sqrt{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }}{\boldsymbol{{y}}}\right] \\ $$
Commented by M±th+et+s last updated on 08/Jun/20
$${thank}\:{you}\:{sir} \\ $$
Commented by smridha last updated on 08/Jun/20
welcome