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Question Number 123598 by aurpeyz last updated on 26/Nov/20

$$provethat\underset{0}{\stackrel{a}{\int }}\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

$${\int }_0^a\sqrt{a^2-x^2}dxx=asin\theta$$$${\int }_0^{\frac{\pi }{2}}acos\theta \sqrt{a^2-a^2{sin}^2\theta }d\theta$$$$=a^2{\int }_0^{\frac{\pi }{2}}{cos}^2\theta =\frac{a^2}{2}{\int }_0^{\frac{\pi }{2}}1+cos2\theta dx$$$$=\frac{\pi a^2}{4}+\frac{a^2}{2}{\int }_0^{\frac{\pi }{2}}cos2\theta =\frac{\pi a^2}{4}+\frac{a^2}{4}{\left[cos2\theta \right]}_0^{\frac{\pi }{2}}=\frac{\pi a^2}{4}$$

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