Question Number 77158 by TawaTawa last updated on 03/Jan/20
$$\int_{−\mathrm{2}} ^{\:\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{3}} \:\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}\:−\:\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{dx} \\ $$
Commented by TawaTawa last updated on 03/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 03/Jan/20
$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi×\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}=\pi \\ $$
Commented by TawaTawa last updated on 03/Jan/20
$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{show}\:\mathrm{more}\:\mathrm{steps}.\:\mathrm{Please} \\ $$
Commented by mr W last updated on 03/Jan/20
$$\int_{−\mathrm{2}} ^{\:\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{3}} \:\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}\:−\:\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{dx} \\ $$$$=\int_{−\mathrm{2}} ^{\:\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{3}} \:\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\:\mathrm{2}} \sqrt{\mathrm{4}\:−\:\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{dx} \\ $$$$=\mathrm{0}+\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\:\mathrm{2}} \sqrt{\mathrm{4}\:−\:\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{dx} \\ $$$$=\mathrm{0}+\:\frac{\mathrm{1}}{\mathrm{2}}×{area}\:{of}\:{semicircle}\:{radius}\:\mathrm{2} \\ $$$$=\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi×\mathrm{2}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\pi \\ $$