Evaluate the integral of (cos⁻¹ x)2 with respect to x.

Let $u = (\cos^{-1} x)^2$ and $dv = dx$.
Then, we find $du$ and $v$:

$$du = 2(\cos^{-1} x) \cdot \frac{d}{dx}(\cos^{-1} x) \, dx = 2(\cos^{-1} x) \left(\frac{-1}{\sqrt{1-x^2}}\right) \, dx$$ $$v = \int dx = x$$

The integration by parts formula is $\int u \, dv = uv - \int v \, du$.

$$\int (\cos^{-1} x)^2 dx = x(\cos^{-1} x)^2 - \int x \cdot 2(\cos^{-1} x) \left(\frac{-1}{\sqrt{1-x^2}}\right) \, dx$$ $$\int (\cos^{-1} x)^2 dx = x(\cos^{-1} x)^2 + 2 \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx$$

Consider the integral $I_2 = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx$.
Let $U = \cos^{-1} x$ and $dV = \frac{x}{\sqrt{1-x^2}} \, dx$.
Then, we find $dU$ and $V$:

$$dU = \frac{-1}{\sqrt{1-x^2}} \, dx$$

To find $V$, integrate $dV$:
Let $w = 1-x^2$, so $dw = -2x \, dx$, which means $x \, dx = -\frac{1}{2} dw$.

$$V = \int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-1/2} \, dw$$ $$V = -\frac{1}{2} \left(\frac{w^{1/2}}{1/2}\right) = -w^{1/2} = -\sqrt{1-x^2}$$

Now apply the integration by parts formula $UV - \int V \, dU$ for $I_2$:

$$I_2 = (\cos^{-1} x)(-\sqrt{1-x^2}) - \int (-\sqrt{1-x^2}) \left(\frac{-1}{\sqrt{1-x^2}}\right) \, dx$$ $$I_2 = -\sqrt{1-x^2} \cos^{-1} x - \int 1 \, dx$$ $$I_2 = -\sqrt{1-x^2} \cos^{-1} x - x$$

$$\int (\cos^{-1} x)^2 dx = x(\cos^{-1} x)^2 + 2 \left( -\sqrt{1-x^2} \cos^{-1} x - x \right) + C$$ $$\int (\cos^{-1} x)^2 dx = x(\cos^{-1} x)^2 - 2\sqrt{1-x^2} \cos^{-1} x - 2x + C$$

The final answer is $x(\cos^{-1} x)^2 - 2\sqrt{1-x^2} \cos^{-1} x - 2x + C$.