Calculate the integral: integral from - (sqrt(2))/2

Calculate the integral: $\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}}$.

We will use the Newton-Leibniz formula after recognizing the antiderivative.

Step 1: Identify the antiderivative

We recognize that:

\[\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\]

Therefore, the antiderivative is:

\[F(x) = \arcsin(x) + C\]

Step 2: Apply the Newton-Leibniz formula

\[\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}} = \left[\arcsin(x)\right]_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}\]

Step 3: Evaluate at the upper limit $x = \frac{\sqrt{2}}{2}$

Recall that $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, therefore:

\[\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\]

Step 4: Evaluate at the lower limit $x = -\frac{\sqrt{2}}{2}$

Since $\arcsin$ is an odd function: $\arcsin(-x) = -\arcsin(x)$:

\[\arcsin\left(-\frac{\sqrt{2}}{2}\right) = -\arcsin\left(\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}\]

Step 5: Calculate the definite integral

\[\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}} = \arcsin\left(\frac{\sqrt{2}}{2}\right) - \arcsin\left(-\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}\]

Alternative approach using symmetry:

Since $f(x) = \frac{1}{\sqrt{1-x^2}}$ is an even function, we can write:

\[\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}} = 2\int_0^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}} = 2\left[\arcsin(x)\right]_0^{\frac{\sqrt{2}}{2}} = 2\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{2}\]

This confirms our result!

Geometric interpretation:

The expression $\sqrt{1-x^2}$ comes from the equation of the unit circle: $x^2+y^2=1$, which gives $y=\sqrt{1-x^2}$ for the upper semicircle. The integrand $\frac{1}{\sqrt{1-x^2}}$ represents the reciprocal of the $y$-coordinate on the unit circle, and the integral computes the arc (measured as an angle in radians) along the unit circle from $x=-\frac{\sqrt2}2$ to $x=\frac{\sqrt2}2$. Since these points correspond to angles $\mp\frac\pi4$, the angular span is $\frac{\pi}{2}$ radians (90°) — a quarter of the full circle.

Connection to the unit circle:

The points $\left(\pm\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ lie on the unit circle at angles of $\pm 45°$ from the positive $x$-axis. The arc between these points (on the upper semicircle) subtends an angle of $90° = \frac{\pi}{2}$ radians.

Final answer:

\[\boxed{\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}}\]