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Question Number 136458 by aurpeyz last updated on 22/Mar/21

$$\int \sqrt{1-4x^2}dx$$

Answered by mathmax by abdo last updated on 22/Mar/21

$$\mathrm{I}=\int \sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{chamgement}2\mathrm{x}=\sin \theta \Rightarrow$$$$\mathrm{I}=\int \cos \theta \frac{\cos \theta }{2}\mathrm{d}\theta =\int \frac{1}{2}{\cos }^2\theta \mathrm{d}\theta =\frac{1}{4}\int (1+\cos \left(2\theta \right)\mathrm{d}\theta$$$$=\frac{\theta }{4}+\frac{1}{8}\sin \left(2\theta \right)+\mathrm{C}=\frac{\theta }{4}+\frac{1}{4}\sin \theta \cos \theta +\mathrm{C}$$$$=\arcsin \left(2\mathrm{x}\right)+\frac{1}{4}\left(2\mathrm{x}\right)\sqrt{1-{4\mathrm{x}}^2}+\mathrm{C}\Rightarrow$$$$\mathrm{I}=\arcsin \left(2\mathrm{x}\right)+\frac{\mathrm{x}}{2}\sqrt{1-{4\mathrm{x}}^2}+\mathrm{C}$$

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