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Question Number 125837 by joki last updated on 14/Dec/20

$$provethat:$$$$\int \sqrt{a^2-x^2}dx{=}_{}\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{sin}^{-1}\left(\frac{x}{a}\right)+c$$

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

$$Takex=asin\theta ,1=acos\theta \frac{d\theta }{dx}$$$$\int \sqrt{a^2-x^2}dx=\int acos\theta \sqrt{a^2-a^2{sin}^2\theta }d\theta$$$$=a^2\int {cos}^2\theta d\theta =\frac{a^2}{2}\int 1+cos2\theta d\theta =\frac{a^2\theta }{2}+\frac{a^2}{4}sin2\theta +C$$$$\theta ={sin}^{-1}\frac{x}{a}sin2\theta =2\frac{x}{a^2}\sqrt{a-x^2}$$$$So\int \sqrt{a^2-x^2}dx=\frac{a^2{sin}^{-1}\frac{x}{a}}{2}+\frac{x}{2}\sqrt{a^2-x^2}+C$$

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