`int_0^1 (cos^-1 x)^2 dx`.

Answer

To solve the integral \( \int_0^1 (\cos^{-1} x)^2 \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( \theta = \cos^{-1} x \). Then, we have:
\[ x = \cos \theta \]
Differentiating both sides gives:
\[ dx = -\sin \theta \, d\theta \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ \theta = \cos^{-1}(0) = \frac{\pi}{2} \]
When \( x = 1 \):
\[ \theta = \cos^{-1}(1) = 0 \]
Thus, the limits change from \( x: 0 \to 1 \) to \( \theta: \frac{\pi}{2} \to 0 \).

### Step 3: Rewrite the integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ \int_0^1 (\cos^{-1} x)^2 \, dx = \int_{\frac{\pi}{2}}^0 \theta^2 (-\sin \theta) \, d\theta \]
This can be rewritten as:
\[ \int_0^{\frac{\pi}{2}} \theta^2 \sin \theta \, d\theta \]

### Step 4: Integration by parts
Let:
- \( u = \theta^2 \) ⇒ \( du = 2\theta \, d\theta \)
- \( dv = \sin \theta \, d\theta \) ⇒ \( v = -\cos \theta \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int_0^{\frac{\pi}{2}} \theta^2 \sin \theta \, d\theta = \left[ -\theta^2 \cos \theta \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 2\theta \cos \theta \, d\theta \]

### Step 5: Evaluate the boundary term
Evaluating the boundary term:
\[ \left[ -\theta^2 \cos \theta \right]_0^{\frac{\pi}{2}} = -\left(\frac{\pi}{2}\right)^2 \cdot \cos\left(\frac{\pi}{2}\right) + 0 = 0 \]

### Step 6: Second integration by parts
Now we need to evaluate:
\[ \int_0^{\frac{\pi}{2}} 2\theta \cos \theta \, d\theta \]
Let:
- \( u = 2\theta \) ⇒ \( du = 2 \, d\theta \)
- \( dv = \cos \theta \, d\theta \) ⇒ \( v = \sin \theta \)

Using integration by parts again:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int_0^{\frac{\pi}{2}} 2\theta \cos \theta \, d\theta = \left[ 2\theta \sin \theta \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} 2 \sin \theta \, d\theta \]

### Step 7: Evaluate the boundary term again
Evaluating the boundary term:
\[ \left[ 2\theta \sin \theta \right]_0^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2}\right) \cdot 1 - 0 = \pi \]

### Step 8: Evaluate the remaining integral
Now we need to evaluate:
\[ \int_0^{\frac{\pi}{2}} 2 \sin \theta \, d\theta = -2 \left[ \cos \theta \right]_0^{\frac{\pi}{2}} = -2(0 - (-1)) = 2 \]

### Step 9: Combine results
Putting it all together:
\[ \int_0^{\frac{\pi}{2}} 2\theta \cos \theta \, d\theta = \pi - 2 \]
Thus:
\[ \int_0^{\frac{\pi}{2}} \theta^2 \sin \theta \, d\theta = 0 + (\pi - 2) = \pi - 2 \]

### Final Answer
The value of the integral \( \int_0^1 (\cos^{-1} x)^2 \, dx \) is:
\[ \boxed{\pi - 2} \]