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Question Number 124446 by ZiYangLee last updated on 03/Dec/20

$${\int }_{-2}^2(x^3\cos \frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx=?$$

Answered by Ar Brandon last updated on 03/Dec/20

$$\mathcal{I}={\int }_{-2}^2({\mathrm{x}}^3\cos \frac{\mathrm{x}}{2}+\frac{1}{2})\sqrt{4-{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{1}{2}{\int }_{-2}^2\sqrt{4-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}{\{\frac{\mathrm{x}\sqrt{4-{\mathrm{x}}^2}}{2}+\frac{4}{2}\mathrm{Arcsin}\left(\frac{\mathrm{x}}{2}\right)\}}_{-2}^2$$$$=\frac{\pi }{2}-(-\frac{\pi }{2})=\pi$$

Commented by Ar Brandon last updated on 03/Dec/20

$${\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\{\begin{array}{l}0\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{odd}\\ 2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{even}\end{array}$$$$\int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$$$$Canbeprovedbysettingx=asin\theta$$

Answered by Dwaipayan Shikari last updated on 03/Dec/20

$${\int }_{-2}^2(x^{3}cos\frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx={\int }_{-2}^2(-x^{3}cos\frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx=I$$$$2I={\int }_{-2}^2\sqrt{4-x^2}dx$$$$2I={\int }_{-2}^2\sqrt{4-x^2}dx={\int }_{-1}^1\sqrt{4-4{sin}^2\theta }d\theta x=2sin\theta \Rightarrow 1=2cos\theta \frac{d\theta }{dx}$$$$=2{\int }_0^{\frac{\pi }{2}}1+cos2\theta =\pi +2{\int }_0^{\frac{\pi }{2}}cos2\theta =\pi +\frac{1}{2}{\left[sin2\theta \right]}_0^{\frac{\pi }{2}}=\pi$$

Commented by Dwaipayan Shikari last updated on 03/Dec/20

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

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