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Question Number 95673 by Rio Michael last updated on 26/May/20

$${\int }_0^1\frac{1}{\sqrt{3+4x-4x^2}}dx=?$$

Commented by Tony Lin last updated on 26/May/20

$${\int }_0^1\frac{1}{\sqrt{3+4x-4x^2}}dx$$$$={\int }_0^1\frac{1}{\sqrt{-4(x^2-x+\frac{1}{4})+4}}dx$$$$=\frac{1}{2}{\int }_0^1\frac{1}{\sqrt{1-{(x-\frac{1}{2})}^2}}dx$$$$={\left[\frac{1}{2}{sin}^{-1}(x-\frac{1}{2})\right]}_0^1$$$$=\frac{\pi }{6}$$

Commented by Rio Michael last updated on 26/May/20

$$\mathrm{thanks}$$

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