Integration Questions

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 bysmridha last updated on 08/Jun/20

welcome