Question

The value of \(\int \cos^{-1}(2x)\,dx\) is given by:

• A. $x \cos^{-1}(2x) - \frac{1}{2} \sqrt{1-4x^2} + C$
• B. $x \sin^{-1}(2x) - \frac{1}{2} \sqrt{1-4x^2} + C$
• C. $x \cos^{-1}(2x) + \frac{1}{2} \sqrt{1-4x^2} + C$
• D. $x \sin^{-1}(2x) + \frac{1}{2} \sqrt{1-4x^2} + C$

Hint:

When integrating any single inverse trigonometric or logarithmic function, always use integration by parts with $dv = dx$ ($v = x$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To integrate inverse trigonometric functions like $\cos^{-1}(2x)$, we use the method of Integration by Parts.
Key Formula or Approach:
The Integration by Parts formula is: \[ \int u \, dv = uv - \int v \, du \] Here, we let: - $u = \cos^{-1}(2x)$
- $dv = dx$

Step 2: Detailed Explanation:
Let us find the differentials: \[ du = \frac{d}{dx} \left( \cos^{-1}(2x) \right) dx = -\frac{2}{\sqrt{1 - (2x)^2}} dx = -\frac{2}{\sqrt{1 - 4x^2}} dx \] \[ v = \int dv = \int dx = x \] Applying the integration by parts formula: \[ \int \cos^{-1}(2x) dx = x \cos^{-1}(2x) - \int x \left( -\frac{2}{\sqrt{1 - 4x^2}} \right) dx \] \[ \int \cos^{-1}(2x) dx = x \cos^{-1}(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}} dx \] Now, we evaluate the remaining integral using substitution: Let $t = 1 - 4x^2$.
Differentiating both sides: \[ dt = -8x \, dx \implies 2x \, dx = -\frac{1}{4} dt \] Substitute these into the integral: \[ \int \frac{2x}{\sqrt{1 - 4x^2}} dx = \int \frac{-\frac{1}{4} dt}{\sqrt{t}} = -\frac{1}{4} \int t^{-1/2} dt \] \[ = -\frac{1}{4} \left( \frac{t^{1/2}}{1/2} \right) = -\frac{1}{2} \sqrt{t} \] Substitute back $t = 1 - 4x^2$: \[ = -\frac{1}{2} \sqrt{1 - 4x^2} \] Combine the terms to get the final solution: \[ \int \cos^{-1}(2x) dx = x \cos^{-1}(2x) - \frac{1}{2} \sqrt{1 - 4x^2} + C \]

Step 3: Final Answer:
The result matches Option (A).