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

We are given the functions \( f(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt \) and \( g(x) = \int_{0}^{\cos^2 x} \sin^{-1} \sqrt{t} \, dt \). We are tasked with finding the value of \( f(x) + g(x) \).
Step 1: Analyze the structure of the integrals.
Both integrals are defined with limits depending on \( \sin^2 x \) and \( \cos^2 x \), respectively. The integrand involves the inverse sine of the square root of \( t \). These functions are symmetric, and we aim to understand the total contribution when both integrals are summed over their respective intervals.

Step 2: Combine the integrals.
Since \( \sin^2 x + \cos^2 x = 1 \), the sum of the two integrals simplifies to an integral over the entire interval from 0 to 1: \[ f(x) + g(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_0^{\cos^2 x} \sin^{-1} \sqrt{t} \, dt = \int_0^1 \sin^{-1} \sqrt{t} \, dt \]
Step 3: Evaluate the combined integral.
We need to evaluate: \[ \int_0^1 \sin^{-1} \sqrt{t} \, dt \] Using the substitution \( \sqrt{t} = \sin \theta \), we convert this into an integral involving \( \theta \), and the final result turns out to be \( \frac{\pi}{4} \).
Step 4: Conclusion.
Thus, the value of \( f(x) + g(x) \) is \( \frac{\pi}{4} \).