Question

If \( f(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt \) and \( g(x) = \int_{0}^{\cos^2 x} \sin^{-1} \sqrt{t} \, dt \), then the value of \( f(x) + g(x) \) is:

• A. \( \pi \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \sin^2 x + \sin x + x \)

Hint:

For integrals involving inverse trigonometric functions, it's helpful to use substitutions to simplify the expression, especially when the integrals involve square roots and trigonometric identities.

Answer 1

The Correct Option is B

The problem defines functions \( f(x) \) and \( g(x) \) as definite integrals. The goal is to determine the value of \( f(x) + g(x) \).
Step 1: Integral Analysis
The integrals' limits depend on \( \sin^2 x \) and \( \cos^2 x \). The integrand includes the inverse sine of the square root of \( t \). Recognizing the symmetry, the approach focuses on the combined contribution of the integrals.
Step 2: Integral Combination
Utilizing \( \sin^2 x + \cos^2 x = 1 \), the sum simplifies to a single integral: \[\nf(x) + g(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_0^{\cos^2 x} \sin^{-1} \sqrt{t} \, dt = \int_0^1 \sin^{-1} \sqrt{t} \, dt \]
Step 3: Combined Integral Evaluation
The integral to evaluate is: \[\n\int_0^1 \sin^{-1} \sqrt{t} \, dt \] Employing the substitution \( \sqrt{t} = \sin \theta \), the integral is transformed and yields a final result of \( \frac{\pi}{4} \).
Step 4: Conclusion
Therefore, \( f(x) + g(x) \) equals \( \frac{\pi}{4} \).