Question Number 117963 by peter frank last updated on 14/Oct/20
Answered by john santu last updated on 14/Oct/20
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} \mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} \:}\:\right){dx}\:=\:\mathrm{0} \\ $$$${then}\:\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}\pi=\pi \\ $$
Answered by Dwaipayan Shikari last updated on 14/Oct/20
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx}={I}=\int_{−\mathrm{2}} ^{\mathrm{2}} \left(−{x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{I}=\int_{−\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\left({Integral}\:{is}\:{symmetric}\right) \\ $$$$\int\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\int\mathrm{2}{cos}\theta\sqrt{\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} \theta}\:{d}\theta=\mathrm{4}\int{cos}^{\mathrm{2}} \theta \\ $$$$=\mathrm{2}\int\mathrm{1}+{cos}\mathrm{2}{x}=\mathrm{2}\theta+{sin}\mathrm{2}\theta=\mathrm{2}{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{sin}\left(\mathrm{2}{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}\right) \\ $$$${So}\int_{−\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\mathrm{2}\pi \\ $$$$\mathrm{2}{I}=\mathrm{2}\pi \\ $$$${I}=\pi \\ $$$$ \\ $$