Find the integral from 0 to 1/4 of 1/square root of 1-4x^2 dx

Question

hartnn · Answer

did you try the substitution,

2x = sin y 
?

hartnn · Answer

then what would be 
1-4x^2 = ... ?

anonymous · Answer

idk

hartnn · Answer

have you solved an integration question before ?

anonymous · Answer

yes, I'm trying to figure out how to get rid of the 4

hartnn · Answer

right, thats was the reason to substitute 
"2"x = sin y 
to get rid of that 4

once you find what $\sqrt {1-4x^2}$ is, you'll come to know

anonymous · Answer

That is a known integral.
$$
\int \frac{1}{\sqrt{1-4 x^2}}
   \, dx=\frac{1}{2} \sin
   ^{-1}(2 x)
$$

anonymous · Answer

$$
\frac{1}{2} \sin
   ^{-1}(2 (1/4))- \frac{1}{2} \sin
   ^{-1}(0)=\frac{1}{2} \sin
   ^{-1}(1/2)=\frac 1 2 \frac \pi 6=\frac \pi {12}
$$

anonymous · Answer

We use the fact that
$$
\frac{d \sin ^{-1}(2
   x)}{
   dx}=\frac{2}{\sqrt{1-4 x^2}}
$$