Question

If \[ f(x)=\int_0^{\sin^2 x}\sin^{-1}\!\sqrt{t}\,dt, \quad g(x)=\int_0^{\cos^2 x}\cos^{-1}\!\sqrt{t}\,dt, \] then the value of \( f(x)+g(x) \) is:

• A. \( \pi \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. depends on \( x \)

Hint:

For paired inverse trig integrals: \begin{itemize} \item Use \( \sin^{-1}u + \cos^{-1}u = \frac{\pi}{2} \). \item Convert limits to match. \end{itemize}

Answer 1

The Correct Option is C

Concept: Use substitution symmetry. Step 1: {\color{red}Let \( t=\sin^2\theta \).} Then integrals mirror each other. Step 2: {\color{red}Use identity.} \[ \sin^{-1}u + \cos^{-1}u = \frac{\pi}{2} \] Apply inside integrals. Step 3: {\color{red}Add integrals.} Combined integral reduces to: \[ \int_0^1 \frac{\pi}{2} \, dt = \frac{\pi}{2} \]