Solve the equation: integral from ln 4 to x

Solve the equation: $$ \int_{\ln 4}^x \frac{dt}{\sqrt{e^t-4}} = \frac{\pi}{4} $$

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

Step 1: Evaluate the integral

We need to find:

$$\int \frac{dt}{\sqrt{e^t-4}}$$

Method: Substitution

Let $u = e^t$, so $du = e^t \, dt$, which gives $dt = \frac{du}{u}$.

The integral becomes:

$$\int \frac{dt}{\sqrt{e^t-4}} = \int \frac{1}{\sqrt{u-4}} \cdot \frac{du}{u} = \int \frac{du}{u\sqrt{u-4}}$$

Now let $w = \sqrt{u-4}$, so $w^2 = u-4$ and $u = w^2 + 4$.

Then $du = 2w \, dw$, and:

$$\int \frac{du}{u\sqrt{u-4}} = \int \frac{2w \, dw}{(w^2+4) \cdot w} = \int \frac{2 \, dw}{w^2+4}$$

$$= 2 \int \frac{dw}{w^2+4} = 2 \cdot \frac{1}{2}\arctan\left(\frac{w}{2}\right) + C = \arctan\left(\frac{w}{2}\right) + C$$

Substituting back $w = \sqrt{u-4} = \sqrt{e^t-4}$:

$$\int \frac{dt}{\sqrt{e^t-4}} = \arctan\left(\frac{\sqrt{e^t-4}}{2}\right) + C$$

Step 2: Verify by differentiation

Let's verify this result:

$$\frac{d}{dt}\left[\arctan\left(\frac{\sqrt{e^t-4}}{2}\right)\right] = \frac{1}{1 + \frac{e^t-4}{4}} \cdot \frac{d}{dt}\left[\frac{\sqrt{e^t-4}}{2}\right]$$

$$= \frac{4}{4 + e^t - 4} \cdot \frac{1}{2} \cdot \frac{e^t}{2\sqrt{e^t-4}} = \frac{4}{e^t} \cdot \frac{e^t}{4\sqrt{e^t-4}} = \frac{1}{\sqrt{e^t-4}}$$

This confirms our antiderivative is correct! ✓

Step 3: Evaluate the definite integral

$$\int_{\ln 4}^x \frac{dt}{\sqrt{e^t-4}} = \left[\arctan\left(\frac{\sqrt{e^t-4}}{2}\right)\right]_{\ln 4}^x$$

At $t = x$:

$$\arctan\left(\frac{\sqrt{e^x-4}}{2}\right)$$

At $t = \ln 4$:

$$e^{\ln 4} = 4 \implies \sqrt{e^{\ln 4}-4} = \sqrt{4-4} = 0$$

Therefore:

$$\arctan\left(\frac{0}{2}\right) = \arctan(0) = 0$$

So:

$$\int_{\ln 4}^x \frac{dt}{\sqrt{e^t-4}} = \arctan\left(\frac{\sqrt{e^x-4}}{2}\right) - 0 = \arctan\left(\frac{\sqrt{e^x-4}}{2}\right)$$

Step 4: Set up and solve the equation

We have:

$$\arctan\left(\frac{\sqrt{e^x-4}}{2}\right) = \frac{\pi}{4}$$

Taking the tangent of both sides:

$$\frac{\sqrt{e^x-4}}{2} = \tan\left(\frac{\pi}{4}\right) = 1$$

$$\sqrt{e^x-4} = 2$$

Squaring both sides:

$$e^x - 4 = 4$$

$$e^x = 8$$

Taking the natural logarithm:

$$x = \ln 8$$

Step 5: Verify the solution

Let's check $x = \ln 8$:

Domain check:

- We need $e^x > 4$ for the integrand to be defined

- $e^{\ln 8} = 8 > 4$ ✓

Verification:

$$\arctan\left(\frac{\sqrt{e^{\ln 8}-4}}{2}\right) = \arctan\left(\frac{\sqrt{8-4}}{2}\right) = \arctan\left(\frac{\sqrt{4}}{2}\right) = \arctan\left(\frac{2}{2}\right) = \arctan(1) = \frac{\pi}{4}$$

This confirms our solution! ✓

Alternative form

Since $8 = 2^3$, we can also write:

$$x = \ln 8 = \ln(2^3) = 3\ln 2$$

Final Answer

$$\boxed{x = \ln 8 = 3\ln 2}$$

Numerically: $x = \ln 8 \approx 2.079$

Geometric interpretation

The integral $\int_{\ln 4}^x \frac{dt}{\sqrt{e^t-4}}$ represents the area under the curve $y = \frac{1}{\sqrt{e^t-4}}$ from $t = \ln 4$ to $t = x$.

Properties of the function $f(t) = \frac{1}{\sqrt{e^t-4}}$:

• Domain: $t > \ln 4$ (the function has a vertical asymptote at $t = \ln 4$)
• The function is positive and decreasing on $(\ln 4, \infty)$
• As $t \to \ln 4^+$, $e^t \to 4^+$, so $f(t) \to +\infty$
• As $t \to \infty$, $e^t \to \infty$, so $f(t) \to 0$

The antiderivative $F(t) = \arctan\left(\frac{\sqrt{e^t-4}}{2}\right)$:

• At $t = \ln 4$: $F(\ln 4) = \arctan(0) = 0$
• At $t = \ln 8$: $F(\ln 8) = \arctan(1) = \frac{\pi}{4}$
• At $t = \ln 20$: $F(\ln 20) = \arctan\left(\frac{\sqrt{16}}{2}\right) = \arctan(2) \approx 1.107$
• As $t \to \infty$: $F(t) \to \arctan(\infty) = \frac{\pi}{2}$

The area from $t = \ln 4$ to $t = \ln 8$ is exactly $\frac{\pi}{4}$.

Connection to standard forms

The substitution technique we used transforms the integral into a standard arctangent form:

$$\int \frac{du}{u\sqrt{u-a}} \quad \text{(where } a = 4 \text{)}$$

This type of integral commonly appears when dealing with exponential functions under radicals, and the arctangent antiderivative is a general pattern to remember.

Summary

1. We used substitution $u = e^t$ followed by $w = \sqrt{u-4}$ to get: $$ \int \frac{dt}{\sqrt{e^t-4}} = \arctan\left(\frac{\sqrt{e^t-4}}{2}\right) + C $$
2. The definite integral is: $$ \int_{\ln 4}^x \frac{dt}{\sqrt{e^t-4}} = \arctan\left(\frac{\sqrt{e^x-4}}{2}\right) $$
3. Setting this equal to $\frac{\pi}{4}$ gives: $$ \frac{\sqrt{e^x-4}}{2} = 1 \implies \sqrt{e^x-4} = 2 \implies e^x = 8 $$
4. Therefore: $x = \ln 8 = 3\ln 2$
5. The solution is verified by direct substitution

Key lessons:

• Double substitution (first $u = e^t$, then $w = \sqrt{u-4}$) can simplify complex integrands
• The arctangent function naturally arises from integrals of the form $\int \frac{dw}{w^2+a^2}$
• Knowing that $\arctan(1) = \frac{\pi}{4}$ is crucial for solving this type of equation
• Always verify that solutions satisfy domain restrictions (here, $x > \ln 4$)
• The lower limit $\ln 4$ creates a vertical asymptote, but the integral converges