To evaluate the integral:

$\int \frac{\left( \cos^{-1}(x) \cdot \sqrt{1 - x^2} \right)^{-1}}{\ln \left( 1 + \frac{\sin \left( 2x \cdot \sqrt{1 - x^2} \right)}{\pi} \right)} \, dx$

we need to examine the components of the integrand carefully. Let's break it down:

1. Numerator Analysis:

   • The term $\cos^{-1}(x)$ is the inverse cosine function, which gives an angle $\theta$ for a given $x = \cos(\theta)$.
   • $\sqrt{1 - x^2}$ is part of the expression that often appears in trigonometric integrals, related to $\sin(\theta)$ when $\theta = \cos^{-1}(x)$.
   • The entire expression $\left( \cos^{-1}(x) \cdot \sqrt{1 - x^2} \right)^{-1}$ represents the reciprocal of this product.

2. Denominator Analysis:

   • The term $\sin \left( 2x \cdot \sqrt{1 - x^2} \right)$ suggests a trigonometric function that might be simplified using identities or substitutions.
   • $\ln \left( 1 + \frac{\sin \left( 2x \cdot \sqrt{1 - x^2} \right)}{\pi} \right)$ appears in the denominator, which complicates direct integration.

Potential Strategy:

Considering the complex form of the integrand, a direct integration might be infeasible. However, some possible approaches include:

• Substitution: Trying a substitution for $\cos^{-1}(x)$ or the expression $\sqrt{1 - x^2}$.
• Trigonometric Identities: Simplifying $\sin(2x \cdot \sqrt{1 - x^2})$ using trigonometric identities.
• Numerical Methods: If a symbolic approach is challenging, a numerical approximation might yield a solution.

Would you like me to explore a specific approach or attempt a numerical solution?

Further Questions to Explore:

1. Would you like a step-by-step guide on trigonometric substitutions?
2. Are you interested in understanding how inverse trigonometric functions impact integrals?
3. Do you want an explanation on handling integrals with logarithmic denominators?
4. Would you like an approximation for this integral over a specific interval?
5. Do you need a breakdown of trigonometric identities that might simplify this expression?

Tip: When handling integrals with inverse trigonometric functions and square roots, substitutions like $x = \sin(\theta)$ or $x = \cos(\theta)$ can often simplify the integrand significantly.