To solve the integral $\int \cos^{-1}\left(\frac{x}{2}\right) dx$ using integration by parts, we need to set up the integral according to the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Let's choose: $u = \cos^{-1}\left(\frac{x}{2}\right) \quad \text{and} \quad dv = dx$

Now, we need to find $du$ and $v$.

1. Differentiate $u$: $u = \cos^{-1}\left(\frac{x}{2}\right)$ $du = \frac{d}{dx} \left( \cos^{-1}\left(\frac{x}{2}\right) \right) dx$

Using the chain rule and the derivative of the inverse cosine function: $\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1 - x^2}}$ we get: $du = \frac{-1}{\sqrt{1 - \left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2} dx$ $du = \frac{-1}{2\sqrt{1 - \frac{x^2}{4}}} dx$ $du = \frac{-1}{2\sqrt{\frac{4 - x^2}{4}}} dx$ $du = \frac{-1}{\sqrt{4 - x^2}} dx$

2. Integrate $dv$: $dv = dx$ $v = x$

Now, applying the integration by parts formula: $\int \cos^{-1}\left(\frac{x}{2}\right) dx = x \cos^{-1}\left(\frac{x}{2}\right) - \int x \left( \frac{-1}{\sqrt{4 - x^2}} \right) dx$ $= x \cos^{-1}\left(\frac{x}{2}\right) + \int \frac{x}{\sqrt{4 - x^2}} dx$

To solve $\int \frac{x}{\sqrt{4 - x^2}} dx$, we can use a substitution. Let $u = 4 - x^2$, then $du = -2x dx$ or $\frac{du}{-2} = x dx$.

So, $\int \frac{x}{\sqrt{4 - x^2}} dx = \int \frac{x}{\sqrt{u}} \cdot \frac{du}{-2x}$ $= -\frac{1}{2} \int \frac{1}{\sqrt{u}} du$ $= -\frac{1}{2} \int u^{-\frac{1}{2}} du$ $= -\frac{1}{2} \cdot 2 u^{\frac{1}{2}}$ $= -\sqrt{u}$ $= -\sqrt{4 - x^2}$

Combining the results: $\int \cos^{-1}\left(\frac{x}{2}\right) dx = x \cos^{-1}\left(\frac{x}{2}\right) - \sqrt{4 - x^2} + C$

So the final answer is: $\int \cos^{-1}\left(\frac{x}{2}\right) dx = x \cos^{-1}\left(\frac{x}{2}\right) - \sqrt{4 - x^2} + C$

Would you like more details or have any questions?

Here are 5 numbered questions you might ask next:

1. How do you apply the substitution method in integrals?
2. Can you explain the steps for integration by parts?
3. What are some common integrals involving inverse trigonometric functions?
4. How do you derive the formulas for the derivatives of inverse trigonometric functions?
5. Can you solve another integral problem for me?

Tip: When using integration by parts, always choose $u$ and $dv$ such that the resulting integral $\int v \, du$ is simpler than the original.