1-4x-2-dx-

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

\int \sqrt{1 - 4 x^{2}} d x

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

I = \int \sqrt{1 - {4 x}^{2}} dx we do the chamgement 2 x = \sin θ \Rightarrow

I = \int \cos θ \frac{\cos θ}{2} d θ = \int \frac{1}{2} \cos^{2} θ d θ = \frac{1}{4} \int (1 + \cos (2 θ) d θ

= \frac{θ}{4} + \frac{1}{8} \sin (2 θ) + C = \frac{θ}{4} + \frac{1}{4} \sin θ \cos θ + C

= \arcsin (2 x) + \frac{1}{4} (2 x) \sqrt{1 - {4 x}^{2}} + C \Rightarrow

I = \arcsin (2 x) + \frac{x}{2} \sqrt{1 - {4 x}^{2}} + C