Solving an Arcsecant Integral
The integral in the image is:
\[
\int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^2}}
\]
This integral is the integral of the arcsecant derivative. We can use the substitution \( x = \sin(\theta) \) to solve it:
\[
dx = \cos(\theta) d\theta
\]
When \( x = 0 \), \( \sin(\theta) = 0 \), so \( \theta = 0 \).
When \( x = \frac{1}{2} \), \( \sin(\theta) = \frac{1}{2} \), which means \( \theta = \frac{\pi}{6} \) since it lies in the first quadrant.
Substituting \( x = \sin(\theta) \) into the integral gives us:
\[
\int_0^{\frac{\pi}{6}} \frac{\cos(\theta)}{\sqrt{1 - \sin^2(\theta)}} d\theta
\]
Now, \( \sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)} = \cos(\theta) \)
So the integral simplifies to:
\[
\int_0^{\frac{\pi}{6}} d\theta
\]
This is just the integral of \(1\) with respect to \( \theta \) from \(0\) to \(\frac{\pi}{6}\), so it equals \( \theta \) evaluated between these two limits:
\[
\theta \Big|_0^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6}
\]
Therefore, the integral equals \( \frac{\pi}{6} \).
