Question

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

Answer 1

did you try the substitution,

2x = sin y
?

Answer 2

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

Answer 3

idk

Answer 4

have you solved an integration question before ?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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