Question

If \(\int e^x \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} \right) dx = g(x) + C\),  where C is the constant of integration, then \(g\left( \frac{1}{2} \right)\)equals:

• A. \( \frac{\pi}{6} \sqrt{\frac{e}{2}} \)
• B. \( \frac{\pi}{4} \sqrt{\frac{e}{2}} \)
• C. \( \frac{\pi}{6} \sqrt{\frac{e}{3}} \)
• D. \( \frac{\pi}{4} \sqrt{\frac{e}{3}} \)

Hint:

When integrating complex expressions involving inverse trigonometric functions: - Apply standard integration formulas for inverse sine and cosine functions. - Recognize patterns in integrals and use substitution to simplify the integral before solving.

Answer 1

The Correct Option is C

To determine \(g\left( \frac{1}{2} \right)\), we must simplify the integral \( \int e^x \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} \right) dx \). The integral consists of three components:

1. \( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \)
2. \( \frac{\sin^{-1} x}{(1-x^2)^{3/2}} \)
3. \( \frac{x}{1-x^2} \)

We observe the pattern \( e^x f(x) \). Applying the Leibniz rule for differentiation of an integral with a parameter, we have:

\(\frac{d}{dx}\left(e^x f(x)\right) = e^x f(x) + e^x f'(x)\)

This suggests the integral can be expressed as:

\(\int e^x f'(x) \, dx \approx e^x f(x)\) (after manipulation)

Given the terms in the integral, we hypothesize:

- Let \( f(x) = \sin^{-1}(x) \) as it appears in the primary terms.

Calculating the derivative of \( f(x) \):

\(\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\), leading to:

\(\int e^x \frac{d}{dx}(\sin^{-1} x) \, dx = e^x \sin^{-1}(x)\)

This result closely matches the structure of the original integral. Therefore, by integration:

\(g(x) = e^x \sin^{-1}(x)\) (determined by inspection and extrapolation of the integral components)

We need to compute \(g\left(\frac{1}{2}\right)\):

Substituting \( x = \frac{1}{2} \) into \(g(x)\):

\(g\left(\frac{1}{2}\right) = e^{\frac{1}{2}} \sin^{-1}\left(\frac{1}{2}\right)\)

Since \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\), we have:

\(g\left(\frac{1}{2}\right) = e^{\frac{1}{2}} \cdot \frac{\pi}{6}\).

Approximating \(e^{\frac{1}{2}}\) as \( \sqrt{e} \):

\(\frac{\sqrt{e} \cdot \pi}{6}\). This is approximately equal to \( \frac{\pi}{4} \sqrt{3} \).

The final answer is \( \frac{\pi}{4} \sqrt{3} \).