Solve the equation: integral from x to ² (t^4-t²)^(-1/2)

Solve the equation: $$ \int_x^2 (t^4-t^2)^{-1/2} \, dt = \frac{\pi}{3} $$

To solve this equation, we need to first evaluate the integral on the left-hand side.

Step 1: Simplify the integrand

$$(t^4-t^2)^{-1/2} = \frac{1}{\sqrt{t^4-t^2}} = \frac{1}{\sqrt{t^2(t^2-1)}} = \frac{1}{|t|\sqrt{t^2-1}}$$

For $t \geq 1$, we have $t > 0$ and $t^2 - 1 > 0$, so:

$$(t^4-t^2)^{-1/2} = \frac{1}{t\sqrt{t^2-1}}$$

Since we're integrating from $x$ to $2$, and we need the integral to exist, we require $x \geq 1$.

Step 2: Evaluate the integral

We need to find:

$$\int \frac{dt}{t\sqrt{t^2-1}}$$

Method: Trigonometric substitution

Let $t = \sec \theta$, so $dt = \sec \theta \tan \theta \, d\theta$ and $\sqrt{t^2-1} = \sqrt{\sec^2 \theta - 1} = |\tan \theta| = \tan \theta$ (for $0 < \theta < \frac{\pi}{2}$).

$$\int \frac{dt}{t\sqrt{t^2-1}} = \int \frac{\sec \theta \tan \theta \, d\theta}{\sec \theta \cdot \tan \theta} = \int d\theta = \theta + C$$

Since $t = \sec \theta$, we have $\theta = \sec^{-1} t = \arccos\left(\frac{1}{t}\right)$.

Therefore:

$$\int \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{t}\right) + C$$

Alternative form:

We can also write this as $\arcsin\left(\frac{1}{t}\right) + C$ using the identity:

$$\arccos u + \arcsin u = \frac{\pi}{2}$$

So $\arccos\left(\frac{1}{t}\right) = \frac{\pi}{2} - \arcsin\left(\frac{1}{t}\right)$, but the simpler form is:

$$\int \frac{dt}{t\sqrt{t^2-1}} = \arcsin\left(\frac{1}{t}\right) + C$$

Let's verify this is correct by differentiating:

$$\frac{d}{dt}\left[\arcsin\left(\frac{1}{t}\right)\right] = \frac{1}{\sqrt{1-\frac{1}{t^2}}} \cdot \left(-\frac{1}{t^2}\right) = \frac{-1}{t^2\sqrt{\frac{t^2-1}{t^2}}} = \frac{-1}{t^2 \cdot \frac{\sqrt{t^2-1}}{|t|}}$$

For $t > 0$:

$$= \frac{-1}{t^2 \cdot \frac{\sqrt{t^2-1}}{t}} = \frac{-1}{t\sqrt{t^2-1}}$$

Hmm, we get a negative sign. Let's reconsider.

Actually, the correct antiderivative is:

$$\int \frac{dt}{t\sqrt{t^2-1}} = -\arcsin\left(\frac{1}{t}\right) + C$$

Or equivalently:

$$\int \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{t}\right) + C$$

Let's verify the $\arccos$ version:

$$\frac{d}{dt}\left[\arccos\left(\frac{1}{t}\right)\right] = \frac{-1}{\sqrt{1-\frac{1}{t^2}}} \cdot \left(-\frac{1}{t^2}\right) = \frac{1}{t^2\sqrt{\frac{t^2-1}{t^2}}} = \frac{1}{t\sqrt{t^2-1}}$$

This is correct! ✓

Step 3: Evaluate the definite integral

$$\int_x^2 \frac{dt}{t\sqrt{t^2-1}} = \left[\arccos\left(\frac{1}{t}\right)\right]_x^2$$

At $t = 2$:

$$\arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$$

At $t = x$:

$$\arccos\left(\frac{1}{x}\right)$$

Therefore:

$$\int_x^2 \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{2}\right) - \arccos\left(\frac{1}{x}\right) = \frac{\pi}{3} - \arccos\left(\frac{1}{x}\right)$$

Step 4: Set up and solve the equation

We have:

$$\frac{\pi}{3} - \arccos\left(\frac{1}{x}\right) = \frac{\pi}{3}$$

Subtracting $\frac{\pi}{3}$ from both sides:

$$-\arccos\left(\frac{1}{x}\right) = 0$$

$$\arccos\left(\frac{1}{x}\right) = 0$$

Since $\arccos(u) = 0$ when $u = 1$:

$$\frac{1}{x} = 1$$

$$x = 1$$

Step 5: Verify the solution

Let's check $x = 1$:

Domain check:

- We need $x \geq 1$ for the integral to be defined ✓

- At $x = 1$, the integrand $\frac{1}{t\sqrt{t^2-1}}$ has a singularity

Let's evaluate the integral as a limit:

$$\int_1^2 \frac{dt}{t\sqrt{t^2-1}} = \lim_{\epsilon \to 0^+} \int_{1+\epsilon}^2 \frac{dt}{t\sqrt{t^2-1}}$$

$$= \lim_{\epsilon \to 0^+} \left[\arccos\left(\frac{1}{t}\right)\right]_{1+\epsilon}^2$$

$$= \lim_{\epsilon \to 0^+} \left[\frac{\pi}{3} - \arccos\left(\frac{1}{1+\epsilon}\right)\right]$$

As $\epsilon \to 0^+$:

$$\frac{1}{1+\epsilon} \to 1 \implies \arccos\left(\frac{1}{1+\epsilon}\right) \to \arccos(1) = 0$$

Therefore:

$$\int_1^2 \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$$

This confirms our solution! ✓

Final Answer

$$\boxed{x = 1}$$

The solution to the equation is $x = 1$.

Geometric interpretation

The integral $\int_x^2 \frac{dt}{t\sqrt{t^2-1}}$ represents the area under the curve $y = \frac{1}{t\sqrt{t^2-1}}$ from $t = x$ to $t = 2$.

Properties of the function $f(t) = \frac{1}{t\sqrt{t^2-1}}$:

• Domain: $t > 1$ (the function has a vertical asymptote at $t = 1$)
• The function is positive and decreasing on $(1, \infty)$
• As $t \to 1^+$, $f(t) \to +\infty$
• As $t \to \infty$, $f(t) \to 0$

The antiderivative $F(t) = \arccos\left(\frac{1}{t}\right)$:

• At $t = 1$: $F(1) = \arccos(1) = 0$
• At $t = 2$: $F(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$
• At $t = \infty$: $F(\infty) = \arccos(0) = \frac{\pi}{2}$

The total area from $t = 1$ to $t = 2$ is exactly $\frac{\pi}{3}$, which is what the equation requires.

Alternative representation using arcsec

Note that since $t = \sec \theta$, we have $\theta = \text{arcsec}(t)$, so:

$$\int \frac{dt}{t\sqrt{t^2-1}} = \text{arcsec}(t) + C$$

This gives:

$$\int_x^2 \frac{dt}{t\sqrt{t^2-1}} = \text{arcsec}(2) - \text{arcsec}(x)$$

Since $\text{arcsec}(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$ and $\text{arcsec}(1) = 0$, we get the same result.

Summary

1. We simplified the integrand: $(t^4-t^2)^{-1/2} = \frac{1}{t\sqrt{t^2-1}}$ for $t \geq 1$
2. We evaluated the integral using the substitution $t = \sec \theta$: $$ \int \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{t}\right) + C $$
3. The definite integral is: $$ \int_x^2 \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{3} - \arccos\left(\frac{1}{x}\right) $$
4. Setting this equal to $\frac{\pi}{3}$ gives $\arccos\left(\frac{1}{x}\right) = 0$, so $x = 1$
5. The solution is verified, noting that the integral is improper at $x = 1$ but converges to $\frac{\pi}{3}$

Key lessons:

• The substitution $t = \sec \theta$ is ideal for integrands involving $\sqrt{t^2-1}$
• The antiderivative $\arccos\left(\frac{1}{t}\right) = \text{arcsec}(t)$ is a useful result to remember
• Improper integrals at endpoints can still yield finite values if they converge
• Always verify solutions, especially when dealing with improper integrals