Step-by-step Solution:

We need to evaluate: \[ I = \int_{0}^{\sin^2 x} \sin^{-1} (5t) \, dt + \int_{0}^{\cos^2 x} \cos^{-1} (5t) \, dt \] We use the property: \[ \sin^{-1} y + \cos^{-1} y = \frac{\pi}{2} \] So, \[ \sin^{-1} (5t) + \cos^{-1} (5t) = \frac{\pi}{2} \] Split the Integral \[ I = \int_{0}^{\sin^2 x} \sin^{-1} (5t) \, dt + \int_{0}^{\cos^2 x} \left( \frac{\pi}{2} - \sin^{-1} (5t) \right) dt \] \[ I = \int_{0}^{\sin^2 x} \sin^{-1} (5t) \, dt + \frac{\pi}{2} \int_{0}^{\cos^2 x} dt - \int_{0}^{\cos^2 x} \sin^{-1} (5t) \, dt \] Since the first and third integrals cancel out, we are left with: \[ I = \frac{\pi}{2} \int_{0}^{\cos^2 x} dt \] \[ = \frac{\pi}{2} \times \cos^2 x \] Since the integral does not yield a fixed value for all \(x\), the answer is None.