Calculate integral from -1 to x (arcsin t)/(sqrt(t+1)) dt

Calculate $\int_{-1}^x \frac{\arcsin t}{\sqrt{t+1}} \, dt$ and then solve the equation: $$ \int_{-1}^x \frac{\arcsin t}{\sqrt{1+t}} = 2\sqrt{x+1} \arcsin x + 4\sqrt{2} $$

Part 1: Calculate the integral

We need to evaluate:

$$I(x) = \int_{-1}^x \frac{\arcsin t}{\sqrt{t+1}} \, dt$$

This integral requires integration by parts. Let's use the substitution approach first to make it easier.

Method: Integration by parts

Let:

$$u = \arcsin t \quad \text{and} \quad dv = \frac{dt}{\sqrt{t+1}}$$

Then:

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

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

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

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

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

Simplifying the remaining integral:

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

$$= \int \frac{1}{\sqrt{1-t}} \, dt$$

Let $u = 1-t$, then $du = -dt$:

$$\int \frac{1}{\sqrt{1-t}} \, dt = -\int \frac{du}{\sqrt{u}} = -2\sqrt{u} = -2\sqrt{1-t}$$

Combining the results:

$$\int \frac{\arcsin t}{\sqrt{t+1}} \, dt = 2\sqrt{t+1} \arcsin t - 2(-2\sqrt{1-t})$$

$$= 2\sqrt{t+1} \arcsin t + 4\sqrt{1-t} + C$$

Evaluating the definite integral:

$$I(x) = \int_{-1}^x \frac{\arcsin t}{\sqrt{t+1}} \, dt = \left[2\sqrt{t+1} \arcsin t + 4\sqrt{1-t}\right]_{-1}^x$$

At $t = x$:

$$2\sqrt{x+1} \arcsin x + 4\sqrt{1-x}$$

At $t = -1$:

$$2\sqrt{-1+1} \arcsin(-1) + 4\sqrt{1-(-1)} = 2 \cdot 0 \cdot \left(-\frac{\pi}{2}\right) + 4\sqrt{2} = 0 + 4\sqrt{2}$$

Therefore:

$$I(x) = 2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2}$$

Answer to Part 1

$$\boxed{\int_{-1}^x \frac{\arcsin t}{\sqrt{t+1}} \, dt = 2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2}}$$

Part 2: Solve the equation

We need to solve:

$$\int_{-1}^x \frac{\arcsin t}{\sqrt{1+t}} = 2\sqrt{x+1} \arcsin x + 4\sqrt{2}$$

Using our result from Part 1:

$$2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2} = 2\sqrt{x+1} \arcsin x + 4\sqrt{2}$$

Simplifying (subtract $2\sqrt{x+1} \arcsin x$ from both sides):

$$4\sqrt{1-x} - 4\sqrt{2} = 4\sqrt{2}$$

$$4\sqrt{1-x} = 8\sqrt{2}$$

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

Squaring both sides:

$$1-x = 8$$

$$x = 1 - 8 = -7$$

Verification

We need to check if $x = -7$ is valid:

Domain check:

• For $\arcsin t$ to be defined, we need $-1 \leq t \leq 1$
• Since we're integrating from $t = -1$ to $t = x$, we need $x \in [-1, 1]$
• But $x = -7 < -1$, which is outside the domain!

This means $x = -7$ is not a valid solution because $\arcsin t$ is not defined for $t < -1$.

Re-examining the problem:

Let's check our algebra. We had:

$$4\sqrt{1-x} - 4\sqrt{2} = 4\sqrt{2}$$

This is correct. But for this equation to make sense, we need:

• $\sqrt{1-x}$ to be defined, so $x \leq 1$
• $\sqrt{1-x} = 2\sqrt{2}$ requires $1-x = 8$, so $x = -7$

However, the integral itself is only defined for $x \in [-1, 1]$ because of the $\arcsin t$ term.

Conclusion:

The equation has no solution in the domain where the integral is defined, i.e., $x \in [-1, 1]$.

Alternatively, if we check what value $\sqrt{1-x}$ can take for $x \in [-1, 1]$:

• When $x = -1$: $\sqrt{1-(-1)} = \sqrt{2}$
• When $x = 1$: $\sqrt{1-1} = 0$
• So for $x \in [-1, 1]$: $\sqrt{1-x} \in [0, \sqrt{2}]$

But we need $\sqrt{1-x} = 2\sqrt{2}$, which is greater than $\sqrt{2}$, so this is impossible for $x \in [-1, 1]$.

Alternative interpretation

Perhaps there's an error in the problem statement, or the equation is meant to illustrate that certain equations have no solutions. Let's verify our calculation of the integral is correct by differentiating.

Verification using the Fundamental Theorem of Calculus:

If $I(x) = 2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2}$, then:

$$I'(x) = \frac{d}{dx}[2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2}]$$

Using the product rule on the first term:

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

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

For the second term:

$$\frac{d}{dx}[4\sqrt{1-x}] = 4 \cdot \frac{-1}{2\sqrt{1-x}} = \frac{-2}{\sqrt{1-x}}$$

Therefore:

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

Simplifying the second and third terms:

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

$$= \frac{2}{\sqrt{1-x}} - \frac{2}{\sqrt{1-x}} = 0$$

So:

$$I'(x) = \frac{\arcsin x}{\sqrt{x+1}}$$

This confirms our integral is correct! ✓

Final Answer

Part 1 - The integral:

$$\boxed{\int_{-1}^x \frac{\arcsin t}{\sqrt{t+1}} \, dt = 2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2}}$$

Part 2 - Solution to the equation:

$$\boxed{\text{No solution in the domain } x \in [-1, 1]}$$

The algebraic solution $x = -7$ is outside the domain where the integral is defined.

Summary

1. We used integration by parts to evaluate the integral: $$ \int \frac{\arcsin t}{\sqrt{t+1}} \, dt = 2\sqrt{t+1} \arcsin t + 4\sqrt{1-t} + C $$
2. The definite integral from $-1$ to $x$ gives: $$ I(x) = 2\sqrt{x+1} \arcsin x + 4\sqrt{1-x} - 4\sqrt{2} $$
3. The equation simplifies to $\sqrt{1-x} = 2\sqrt{2}$, giving $x = -7$
4. However, $x = -7$ is outside the domain $[-1, 1]$ where $\arcsin t$ is defined
5. Therefore, the equation has no solution in the valid domain

Key lesson: When solving equations involving definite integrals, always check that solutions lie within the domain where the integral is defined. Domain restrictions from inverse trigonometric functions are particularly important to verify.