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2-2-x-3-cos-x-2-4-x-2-dx-




Question Number 160969 by mkam last updated on 10/Dec/21
∫_(−2) ^( 2) x^3 cos((x/2)) (√(4−x^2 )) dx
$$\int_{−\mathrm{2}} ^{\:\mathrm{2}} {x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$
Answered by MJS_new last updated on 10/Dec/21
g(x)=f_1 (x)×f_2 (x)×f_3 (x)  f_1 (x)=x^3 ; f_1 (−x)=−f_1 (x)  f_2 (3)=cos (x/2); f_2 (−x)=f_2 (x)  f_3 (x)=(√(4−x^2 )); f_3 (−x)=f_3 (x)  ⇒ g(−x)=−g(x)  ⇒ ∫_(−a) ^a g(x)dx=0
$${g}\left({x}\right)={f}_{\mathrm{1}} \left({x}\right)×{f}_{\mathrm{2}} \left({x}\right)×{f}_{\mathrm{3}} \left({x}\right) \\ $$$${f}_{\mathrm{1}} \left({x}\right)={x}^{\mathrm{3}} ;\:{f}_{\mathrm{1}} \left(−{x}\right)=−{f}_{\mathrm{1}} \left({x}\right) \\ $$$${f}_{\mathrm{2}} \left(\mathrm{3}\right)=\mathrm{cos}\:\frac{{x}}{\mathrm{2}};\:{f}_{\mathrm{2}} \left(−{x}\right)={f}_{\mathrm{2}} \left({x}\right) \\ $$$${f}_{\mathrm{3}} \left({x}\right)=\sqrt{\mathrm{4}−{x}^{\mathrm{2}} };\:{f}_{\mathrm{3}} \left(−{x}\right)={f}_{\mathrm{3}} \left({x}\right) \\ $$$$\Rightarrow\:{g}\left(−{x}\right)=−{g}\left({x}\right) \\ $$$$\Rightarrow\:\underset{−{a}} {\overset{{a}} {\int}}{g}\left({x}\right){dx}=\mathrm{0} \\ $$

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