Question

\( \int \frac{\sin^{-1}x}{\sqrt{1-x^2}} \, dx = \)

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

Hint:

If derivative of inverse trig appears, use substitution directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This integral can be solved using the method of substitution. We look for a part of the integrand whose derivative is also present. This pattern suggests a substitution that will simplify the integral.
Step 2: Key Formula or Approach:
1. Recognize that the derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1-x^2}}\). 2. Use the substitution \(u = \sin^{-1}x\). 3. This implies \(du = \frac{1}{\sqrt{1-x^2}} dx\). 4. Substitute `u` and `du` into the integral and evaluate the new, simpler integral. 5. Substitute back to express the answer in terms of `x`.
Step 3: Detailed Explanation:
The integral is: \[ \int \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx \] We can rewrite this as: \[ \int (\sin^{-1}x) \cdot \left(\frac{1}{\sqrt{1-x^2}}\right) dx \] Let's make the substitution: \[ u = \sin^{-1}x \] Then, the differential `du` is: \[ du = \frac{1}{\sqrt{1-x^2}} dx \] Now we can substitute `u` and `du` into the original integral: \[ \int u \cdot du \] This is a simple power rule integration: \[ \frac{u^2}{2} + C \] Finally, substitute back \(u = \sin^{-1}x\): \[ \frac{(\sin^{-1}x)^2}{2} + C \] Step 4: Final Answer:
The integral is \(\frac{1}{2}(\sin^{-1}x)^2 + C\).