Question

$$\int \cos^{-1} x \, dx =$$

• A. $x \cos^{-1} x + \sqrt{1-x^2} + c$
• B. $-x \cos^{-1} x + \sqrt{1+x^2} + c$
• C. $x \cos^{-1} x - \sqrt{1+x^2} + c$
• D. $x \cos^{-1} x - \sqrt{1-x^2} + c$

Hint:

If you forget the integration steps during an exam, you can differentiate the answer choices instead! Differentiating option (D) using the product rule gives: $\frac{d}{dx}(x\cos^{-1}x - \sqrt{1-x^2}) = (1 \cdot \cos^{-1}x + x \cdot \frac{-1}{\sqrt{1-x^2}}) - \frac{-2x}{2\sqrt{1-x^2}} = \cos^{-1}x$. This confirms choice (D) perfectly in a few seconds!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks us to find the general antiderivative function of the inverse cosine function, $\int \cos^{-1} x \, dx$.

Step 2: Key Formula or Approach:
Since we cannot integrate an inverse function directly, we use the method of Integration by Parts (ILATE rule). We introduce an implicit factor of 1 as our second function: $$\int f(x)g(x) \, dx = f(x)\int g(x) \, dx - \int \left( f'(x) \int g(x) \, dx \right) dx$$ Here, let $f(x) = \cos^{-1} x$ (First function) and $g(x) = 1$ (Second function). Recall that $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$.

Step 3: Detailed Explanation:
Let's apply the integration by parts formula: $$\int (\cos^{-1} x \cdot 1) \, dx = \cos^{-1} x \cdot \int 1 \, dx - \int \left( \frac{d}{dx}(\cos^{-1} x) \cdot \int 1 \, dx \right) dx$$ $$\int \cos^{-1} x \, dx = x \cos^{-1} x - \int \left( -\frac{1}{\sqrt{1-x^2}} \cdot x \right) dx$$ $$\int \cos^{-1} x \, dx = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx \quad \text{--- (Equation 1)}$$ To solve the remaining integral term, use u-substitution. Let $t = 1 - x^2$. Differentiating both sides gives: $$dt = -2x \, dx \implies x \, dx = -\frac{1}{2} dt$$ Substitute these into the integral term from Equation 1: $$\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-\frac{1}{2}} \, dt$$ Using the power rule: $$-\frac{1}{2} \cdot \left( \frac{t^{\frac{1}{2}}}{\frac{1}{2}} \right) = -t^{\frac{1}{2}} = -\sqrt{t}$$ Substitute back $t = 1 - x^2$: $$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2}$$ Now, substitute this result back into Equation 1 to find the complete solution: $$\int \cos^{-1} x \, dx = x \cos^{-1} x - \sqrt{1-x^2} + c$$

Step 4: Final Answer:
The indefinite integral evaluates to $x \cos^{-1} x - \sqrt{1-x^2} + c$, which matches option (D).