Question

How do I find the antiderivative of dx / (sqrt 1-4x^2) ?

Answer 1

Use a substitution, 2x=sin(u), say.

Answer 2

Then\[2dx=\cos u du \]and so\[\int\limits_{}{}\frac{dx}{\sqrt{1-4x^2}}=\frac{1}{2}\int\limits_{}{}\frac{\cos u}{\sqrt{1-\sin^2 u}} du=\frac{1}{2}\int\limits_{}{} du=\frac{u}{2}+c\]

Answer 3

Then \[u=\sin^{-1} \frac{x}{2}\]

Answer 4

So the integral is\[\frac{1}{2}\sin^{-1} \frac{x}{2}+ c\]

Answer 5

sorry, \[\sin^{-1} 2x\] I mean

Answer 6

\[\int\limits_{}{}\frac{dx}{\sqrt{1-4x^2}}=\frac{1}{2}\sin^{-1} 2x +c\]

Answer 7

or the other easy way is to seprate deduction like:\[1\div \sqrt{1-4*x^2}=(A/\sqrt{1-2*x})+(B/\sqrt{1+2x})\]
which A=1/sqrt{2}=B

Answer 8

ijadi, your method's nice

Answer 9

thanks!

Answer 10

thanks!