Calculate the improper integral: integral from 0 to 1

Calculate the improper integral: $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} $$

This is an improper integral because the integrand has a vertical asymptote at $x = 1$.

Step 1: Identify the singularity

As $x \to 1^-$:

$$1 - x^2 \to 0^+ \implies \frac{1}{\sqrt{1-x^2}} \to +\infty$$

Therefore, we must evaluate this as an improper integral.

Step 2: Express as a limit

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}} = \lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{dx}{\sqrt{1-x^2}}$$

Step 3: Recognize the standard antiderivative

The integral $\int \frac{dx}{\sqrt{1-x^2}}$ is a standard form:

$$\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$$

Verification:

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

This confirms our antiderivative is correct! ✓

Step 4: Evaluate the definite integral

$$\int_0^{1-\varepsilon} \frac{dx}{\sqrt{1-x^2}} = [\arcsin x]_0^{1-\varepsilon} = \arcsin(1-\varepsilon) - \arcsin 0$$

Since $\arcsin 0 = 0$:

$$= \arcsin(1-\varepsilon)$$

Step 5: Take the limit

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}} = \lim_{\varepsilon \to 0^+} \arcsin(1-\varepsilon)$$

As $\varepsilon \to 0^+$, we have $(1-\varepsilon) \to 1^-$, so:

$$\lim_{\varepsilon \to 0^+} \arcsin(1-\varepsilon) = \arcsin 1 = \frac{\pi}{2}$$

Final Answer

$$\boxed{\int_0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}}$$

The integral converges to $\frac{\pi}{2}$.

Alternative approach: Direct evaluation

We can also write this directly:

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}} = [\arcsin x]_0^1 = \arcsin 1 - \arcsin 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

This works because the singularity at $x = 1$ is integrable (the integral converges).

Why the integral converges

Near $x = 1$, let $u = 1 - x$, so as $x \to 1^-$, we have $u \to 0^+$.

$$1 - x^2 = (1-x)(1+x) = u(2-u) \approx 2u \quad \text{for small } u$$

So near $x = 1$:

$$\frac{1}{\sqrt{1-x^2}} \approx \frac{1}{\sqrt{2u}} = \frac{1}{\sqrt{2(1-x)}}$$

The integral becomes:

$$\int \frac{dx}{\sqrt{1-x}} \sim \int (1-x)^{-1/2} \, dx$$

This converges because the exponent $-\frac{1}{2}$ satisfies $-\frac{1}{2} > -1$ (i.e., $\alpha = \frac{1}{2} < 1$).

Geometric interpretation

The integral $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$ has a beautiful geometric interpretation.

Connection to the unit circle:

Consider the equation $x^2 + y^2 = 1$ (unit circle). Solving for $y$:

$$y = \sqrt{1-x^2}$$

The derivative is:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$$

The arc length of the curve from $(0, 1)$ to $(x, \sqrt{1-x^2})$ is:

$$s = \int_0^x \sqrt{1 + \left(\frac{dy}{dt}\right)^2} \, dt = \int_0^x \sqrt{1 + \frac{t^2}{1-t^2}} \, dt = \int_0^x \frac{dt}{\sqrt{1-t^2}}$$

So $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$ represents the arc length along the unit circle from $(0, 1)$ to $(1, 0)$, which is exactly $\frac{\pi}{2}$ (a quarter of the circumference).

Connection to inverse trigonometric functions

The function $\arcsin x$ (inverse sine) is defined such that:

$$y = \arcsin x \iff \sin y = x \quad \text{and} \quad -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Key values:

• $\arcsin 0 = 0$
• $\arcsin \frac{1}{2} = \frac{\pi}{6}$
• $\arcsin \frac{\sqrt{2}}{2} = \frac{\pi}{4}$
• $\arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3}$
• $\arcsin 1 = \frac{\pi}{2}$

The integral we computed shows that:

$$\arcsin 1 = \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}$$

This provides a definition of $\arcsin$ through integration.

Related integrals

General form:

$$\int_0^a \frac{dx}{\sqrt{a^2-x^2}} = \arcsin\frac{x}{a} \bigg|_0^a = \arcsin 1 - \arcsin 0 = \frac{\pi}{2}$$

So for any $a > 0$:

$$\int_0^a \frac{dx}{\sqrt{a^2-x^2}} = \frac{\pi}{2}$$

Example:

$$\int_0^5 \frac{dx}{\sqrt{25-x^2}} = \frac{\pi}{2}$$

Symmetric version:

$$\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}} = 2 \int_0^1 \frac{dx}{\sqrt{1-x^2}} = 2 \cdot \frac{\pi}{2} = \pi$$

This is the full semicircle arc length.

Connection to elliptic integrals

When we generalize to:

$$\int_0^1 \frac{dx}{\sqrt{1-k^2x^2}}$$

for $k \neq 1$, we get:

$$\int_0^1 \frac{dx}{\sqrt{1-k^2x^2}} = \frac{1}{k}\arcsin k = \frac{\arcsin k}{k}$$

But when $k = 1$, we get our result: $\frac{\pi}{2}$.

For more complex forms like $\int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$, we enter the realm of elliptic integrals, which cannot be expressed in elementary functions.

Numerical verification

Let's verify numerically by computing $\int_0^{0.99} \frac{dx}{\sqrt{1-x^2}}$:

$$[\arcsin x]_0^{0.99} = \arcsin(0.99) - 0 \approx 1.4292 \text{ radians}$$

And $\frac{\pi}{2} \approx 1.5708$ radians.

The difference is about $0.14$, which comes from the last bit from $0.99$ to $1$.

As we approach 1, the value approaches $\frac{\pi}{2}$.

Summary

1. The integral $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$ has a singularity at $x = 1$
2. The antiderivative is $\arcsin x$
3. The integral converges because near $x = 1$, it behaves like $\int \frac{dx}{\sqrt{1-x}}$, which converges (exponent $\frac{1}{2} < 1$)
4. Direct evaluation gives: $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \arcsin 1 - \arcsin 0 = \frac{\pi}{2} $$
5. Geometrically, this represents a quarter-circle arc length
6. This is a fundamental result connecting integration and inverse trigonometric functions

Key lessons:

• Standard inverse trigonometric antiderivatives are essential
• Not all singularities cause divergence
• The integral $\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x$ is fundamental
• Geometric interpretations provide insight into integral values
• The appearance of $\pi$ connects to circular arc length
• This integral is one of the most important in calculus