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Question Number 117963 by peter frank last updated on 14/Oct/20

Answered by john santu last updated on 14/Oct/20

$${\int }_{-2}^2\left(x^3\cos \left(\frac{x}{2}\right)\sqrt{4-x^2}\right)dx=0$$$$then{\int }_{-2}^2\frac{1}{2}\sqrt{4-x^2}dx={\int }_0^2\sqrt{4-x^2}dx$$$$=\frac{1}{4}\times 4\pi =\pi$$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

$${\int }_{-2}^2(x^{3}cos\frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx=I={\int }_{-2}^2(-x^{3}cos\frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}$$$$2I={\int }_{-2}^2\sqrt{4-x^2}\left(Integralissymmetric\right)$$$$\int \sqrt{4-x^2}dx=\int 2cos\theta \sqrt{4-4{sin}^2\theta }d\theta =4\int {cos}^2\theta$$$$=2\int 1+cos2x=2\theta +sin2\theta =2{sin}^{-1}\frac{x}{2}+sin\left(2{sin}^{-1}\frac{x}{2}\right)$$$$So{\int }_{-2}^2\sqrt{4-x^2}dx=2\pi$$$$2I=2\pi$$$$I=\pi$$$$$$

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