Integration Questions

Question Number 77087 by Dah Solu Tion last updated on 03/Jan/20

$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{{a}}}{\mathrm{2}}} \boldsymbol{{x}}^{\mathrm{2}} \left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} \right)^{\frac{−\mathrm{3}}{\mathrm{2}}} \boldsymbol{{dx}} \\$$$$\boldsymbol{{Help}}!!! \\$$$$\\$$

Commented by turbo msup by abdo last updated on 03/Jan/20

$${changement}\:{x}={asint}\:{give} \\$$$$\int_{\mathrm{0}} ^{\frac{{a}}{\mathrm{2}}} \:{x}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:{dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:{a}^{\mathrm{2}} \:{sin}^{\mathrm{2}} {t}\:×{a}^{−\mathrm{3}} \left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:{acos}\:{t}\:{dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:{sin}^{\mathrm{2}} {t}×\left({cost}\right)^{−\mathrm{3}} \:{cost}\:{dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:{tan}^{\mathrm{2}} {t}\:{dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}−\mathrm{1}\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}−\frac{\pi}{\mathrm{6}} \\$$$$=\left[{tant}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} −\frac{\pi}{\mathrm{6}}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\frac{\pi}{\mathrm{6}} \\$$

Answered by john santu last updated on 03/Jan/20

$$\underset{\mathrm{0}\:\:} {\overset{\frac{{a}}{\mathrm{2}}} {\int}}\:\frac{{x}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}\: \\$$$${let}\:{x}\:=\:{a}\:\mathrm{sin}\:{t}\:\Rightarrow\:{dx}\:=\:{a}\mathrm{cos}\:{t}\:{dt}\: \\$$$${I}\:=\:\int\:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}\:.\:{a}\mathrm{cos}\:{t}\:{dt}}{{a}^{\mathrm{3}} \:\mathrm{cos}\:^{\mathrm{3}} {t}} \\$$$$=\:\int\:\mathrm{tan}\:^{\mathrm{2}} {t}\:{dt}\:=\:\int\:\mathrm{sec}\:^{\mathrm{2}} {t}−\mathrm{1}\:{dt} \\$$$$=\:\mathrm{tan}\:{t}\:−\:{t}\:=\:\frac{{x}}{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\underset{\mathrm{0}} {\overset{\frac{{a}}{\mathrm{2}}} {\mid}} \\$$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\frac{\pi}{\mathrm{6}}\bigstar \\$$