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Question Number 126133 by I want to learn more last updated on 17/Dec/20

Answered by Dwaipayan Shikari last updated on 17/Dec/20

$${\int }_{-2}^2(x^{3}cos\frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx={\int }_{-2}^2-x^{3}cos\frac{x}{2}+\frac{1}{2}\sqrt{4-x^2}dx=I$$$$2I={\int }_{-2}^2\sqrt{4-x^2}dxx=2u$$$$I=2{\int }_{-1}^1\sqrt{1-u^2}du=4{\int }_0^1{(1-u^2)}^{\frac{1}{2}}du{u}^2=t\Rightarrow 2u=\frac{dt}{du}$$$$=2{\int }_0^1{t}^{-\frac{1}{2}}{(1-t)}^{\frac{1}{2}}dt=2\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)=2.\sqrt{\pi }.\frac{\sqrt{\pi }}{2}=\pi$$

Commented by I want to learn more last updated on 17/Dec/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{but}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{the}\mathrm{steps}.\mathrm{the}\mathrm{first}\mathrm{step}\mathrm{precisely}.$$

Commented by Dwaipayan Shikari last updated on 17/Dec/20

$$Theintegralissymmetric$$$${\int }_{-2}^2{x}^{3}cos\frac{x}{2}\sqrt{4-x^2}={\int }_{-2}^2(-x^{3}cos(-\frac{x}{2})\sqrt{4-x^2}dx)=I$$$$2I=0\Rightarrow I=0$$

Commented by I want to learn more last updated on 17/Dec/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

Commented by mathmax by abdo last updated on 17/Dec/20

$$\mathrm{not}\mathrm{correct}!$$

Answered by mathmax by abdo last updated on 17/Dec/20

$$\mathrm{I}={\int }_{-2}^2({\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)+\frac{1}{2})\sqrt{4-{\mathrm{x}}^2}\mathrm{dx}\Rightarrow$$$$\mathrm{I}={\int }_{-2}^2{\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}+\frac{1}{2}{\int }_{-2}^2\sqrt{4-{\mathrm{x}}^2}\mathrm{dx}\mathrm{but}$$$${\int }_{-2}^2{\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}=0\left(\mathrm{function}\mathrm{under}\mathrm{integral}\mathrm{is}\mathrm{odd}\right)$$$$\Rightarrow \mathrm{I}={\int }_0^2\sqrt{4-{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=2\sin \theta }{\int }_0^{\frac{\pi }{2}}2\cos \theta \left(2\cos \theta \right)\mathrm{d}\theta$$$$=4{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta =4{\int }_0^{\frac{\pi }{2}}\frac{1+\cos \left(2\theta \right)}{2}\mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}(1+\cos \left(2\theta \right)\mathrm{d}\theta$$$$=2{[\theta +\frac{1}{2}\sin \left(2\theta \right)]}_0^{\frac{\pi }{2}}=2.\frac{\pi }{2}=\pi$$

Commented by I want to learn more last updated on 20/Dec/20

$$\mathrm{Thanks}\mathrm{sir}.$$

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