To solve this integral, let’s break down the steps for evaluating $\int \cos^{-1}(x) \, dx$.

Solution Outline

The integral is: $\int \cos^{-1}(x) \, dx$

1. Apply Integration by Parts: We can use integration by parts, which states: $\int u \, dv = uv - \int v \, du$ Here, we can set:

   • $u = \cos^{-1}(x)$ (so that $du = -\frac{1}{\sqrt{1 - x^2}} \, dx$)
   • $dv = dx$ (which implies $v = x$)

2. Compute Each Component:

   • $u = \cos^{-1}(x) \Rightarrow du = -\frac{1}{\sqrt{1 - x^2}} \, dx$
   • $dv = dx \Rightarrow v = x$

3. Apply the Integration by Parts Formula: Substitute into the integration by parts formula: $\int \cos^{-1}(x) \, dx = x \cos^{-1}(x) - \int x \left(-\frac{1}{\sqrt{1 - x^2}}\right) \, dx$ Simplify the integral: $= x \cos^{-1}(x) + \int \frac{x}{\sqrt{1 - x^2}} \, dx$

4. Solve the Remaining Integral: Notice that: $\int \frac{x}{\sqrt{1 - x^2}} \, dx$ can be solved by a simple substitution $u = 1 - x^2$, which gives: $\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\sqrt{1 - x^2}$

5. Combine Results: Substituting back, we have: $\int \cos^{-1}(x) \, dx = x \cos^{-1}(x) + \sqrt{1 - x^2} + C$

Final Answer

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

Thus, the answer matches option C.

Would you like further details on each step?

Related Questions

1. How do you apply integration by parts to other inverse trigonometric functions?
2. What is the substitution method, and when is it useful in integration?
3. How does one differentiate and integrate inverse trigonometric functions?
4. How do you handle integrals with trigonometric substitutions?
5. What are some common integrals involving inverse trigonometric functions?

Tip

When dealing with integrals involving inverse trigonometric functions, consider using integration by parts if you see a product of functions (like $x \cos^{-1}(x)$).