Solve the equation: integral from 0 to x (t arctan

Solve the equation: $$ \int_0^x \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \sqrt{1+x^2} \arctan x $$

To solve this equation, we need to first calculate the integral on the left-hand side, then set it equal to the right-hand side and solve for $x$.

Step 1: Calculate the integral

We need to evaluate:

$$I(x) = \int_0^x \frac{t \arctan t}{\sqrt{1+t^2}} \, dt$$

Method: Integration by parts

Let:

$$u = \arctan t \quad \text{and} \quad dv = \frac{t \, dt}{\sqrt{1+t^2}}$$

Then:

$$du = \frac{dt}{1+t^2} \quad \text{and} \quad v = \int \frac{t \, dt}{\sqrt{1+t^2}}$$

To find $v$, let $w = 1+t^2$, so $dw = 2t \, dt$:

$$v = \int \frac{t \, dt}{\sqrt{1+t^2}} = \frac{1}{2}\int \frac{dw}{\sqrt{w}} = \frac{1}{2} \cdot 2\sqrt{w} = \sqrt{1+t^2}$$

Using integration by parts $\int u \, dv = uv - \int v \, du$:

$$\int \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \sqrt{1+t^2} \arctan t - \int \sqrt{1+t^2} \cdot \frac{dt}{1+t^2}$$

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

Now we need to evaluate $\int \frac{dt}{\sqrt{1+t^2}}$.

Evaluating $\int \frac{dt}{\sqrt{1+t^2}}$:

This is a standard integral. Using the substitution $t = \sinh u$ (or recognizing the standard form):

$$\int \frac{dt}{\sqrt{1+t^2}} = \ln|t + \sqrt{1+t^2}| + C = \sinh^{-1}(t) + C$$

Since $t + \sqrt{1+t^2} > 0$ for all real $t$, we can drop the absolute value:

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

Combining the results:

$$\int \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \sqrt{1+t^2} \arctan t - \ln(t + \sqrt{1+t^2}) + C$$

Step 2: Evaluate the definite integral

$$I(x) = \int_0^x \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \left[\sqrt{1+t^2} \arctan t - \ln(t + \sqrt{1+t^2})\right]_0^x$$

At $t = x$:

$$\sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2})$$

At $t = 0$:

$$\sqrt{1+0^2} \arctan 0 - \ln(0 + \sqrt{1+0^2}) = 1 \cdot 0 - \ln(1) = 0 - 0 = 0$$

Therefore:

$$I(x) = \sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2})$$

Step 3: Verify using the Fundamental Theorem of Calculus

Let's verify by differentiating $I(x)$:

$$I'(x) = \frac{d}{dx}\left[\sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2})\right]$$

For the first term (product rule):

$$\frac{d}{dx}[\sqrt{1+x^2} \arctan x] = \frac{x}{\sqrt{1+x^2}} \arctan x + \sqrt{1+x^2} \cdot \frac{1}{1+x^2}$$

$$= \frac{x \arctan x}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}}$$

For the second term (chain rule):

$$\frac{d}{dx}[\ln(x + \sqrt{1+x^2})] = \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}}\right)$$

$$= \frac{1}{x + \sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}$$

Therefore:

$$I'(x) = \frac{x \arctan x}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}} - \frac{1}{\sqrt{1+x^2}} = \frac{x \arctan x}{\sqrt{1+x^2}}$$

This confirms our integral is correct! ✓

Step 4: Set up the equation

We have:

$$\sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2}) = \sqrt{1+x^2} \arctan x$$

Step 5: Solve for $x$

Subtracting $\sqrt{1+x^2} \arctan x$ from both sides:

$$-\ln(x + \sqrt{1+x^2}) = 0$$

$$\ln(x + \sqrt{1+x^2}) = 0$$

Taking the exponential of both sides:

$$x + \sqrt{1+x^2} = e^0 = 1$$

Solving for $\sqrt{1+x^2}$:

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

Domain consideration:

For $\sqrt{1+x^2} = 1-x$ to be valid, we need $1-x \geq 0$, so $x \leq 1$.

Squaring both sides (valid when $x \leq 1$):

$$1 + x^2 = (1-x)^2$$

$$1 + x^2 = 1 - 2x + x^2$$

$$0 = -2x$$

$$x = 0$$

Step 6: Verify the solution

Let's check $x = 0$:

Left-hand side:

$$\int_0^0 \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = 0$$

Right-hand side:

$$\sqrt{1+0^2} \arctan 0 = 1 \cdot 0 = 0$$

Both sides equal 0. ✓

Let's also verify that our squaring step didn't introduce extraneous solutions:

- At $x = 0$: $\sqrt{1+0^2} = 1$ and $1-0 = 1$ ✓

- The condition $x \leq 1$ is satisfied ✓

Final Answer

$$\boxed{x = 0}$$

The unique solution to the equation is $x = 0$.

Geometric interpretation

The equation states that the area under the curve $\frac{t \arctan t}{\sqrt{1+t^2}}$ from $0$ to $x$ equals $\sqrt{1+x^2} \arctan x$.

Our calculation shows that:

$$\int_0^x \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2})$$

For this to equal $\sqrt{1+x^2} \arctan x$, we need the "correction term" $\ln(x + \sqrt{1+x^2})$ to equal zero, which happens only at $x = 0$.

Behavior of the correction term:

The function $g(x) = \ln(x + \sqrt{1+x^2})$ has the following properties:

• $g(0) = \ln(0 + 1) = 0$ (the solution point)
• $g(x) > 0$ for $x > 0$ (since $x + \sqrt{1+x^2} > 1$)
• $g(x) < 0$ for $x < 0$ (since $x + \sqrt{1+x^2} < 1$ when $x < 0$)
• $g(x) = \sinh^{-1}(x)$ (inverse hyperbolic sine)

This is consistent with our finding that $x = 0$ is the unique solution.

Summary

1. We evaluated the integral using integration by parts: $$ \int_0^x \frac{t \arctan t}{\sqrt{1+t^2}} \, dt = \sqrt{1+x^2} \arctan x - \ln(x + \sqrt{1+x^2}) $$
2. Setting this equal to $\sqrt{1+x^2} \arctan x$ gives: $$ \ln(x + \sqrt{1+x^2}) = 0 $$
3. This simplifies to $x + \sqrt{1+x^2} = 1$, which has the unique solution $x = 0$
4. The solution is verified by direct substitution

Key lessons:

• Integration by parts is essential for integrals involving products of algebraic and inverse trigonometric functions
• When squaring equations, always check for extraneous solutions and domain restrictions
• The function $\ln(x + \sqrt{1+x^2}) = \sinh^{-1}(x)$ is an important inverse hyperbolic function that appears naturally in integration
• Verification of solutions by substitution is crucial, especially when algebraic manipulations like squaring are involved