Question:
\(y = \int_{1/8}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{1/8}^{\cos^2 x} \cos^{-1}\sqrt{t} dt, 0 \leq x \leq \pi/2\)
\(y = \int_{1/8}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{1/8}^{\cos^2 x} \cos^{-1}\sqrt{t} dt, 0 \leq x \leq \pi/2\)
Show Hint
Use Leibniz rule and simplify using the given interval.
Updated On: Apr 23, 2026
- Is the equation of a straight line parallel to the \(x\)-axis
- Is the equation of a straight line which is the bisector of first quadrant
- Is the equation of a straight line which is the bisector of second quadrant
- None of the above
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \frac{dy}{dx} = \sin^{-1}(\sin x) \cdot 2\sin x\cos x + \cos^{-1}(\cos x) \cdot (-\sin 2x) \]
Step 2: Calculation / Simplification}
For \(0 \leq x \leq \pi/2\): \(\sin^{-1}(\sin x) = x\), \(\cos^{-1}(\cos x) = x\)
\(\frac{dy}{dx} = x \cdot \sin 2x - x \cdot \sin 2x = 0\)
\(\therefore y\) is constant (parallel to \(x\)-axis).
Step 3: Final Answer
\[ \text{Straight line parallel to the } x\text{-axis} \]
\[ \frac{dy}{dx} = \sin^{-1}(\sin x) \cdot 2\sin x\cos x + \cos^{-1}(\cos x) \cdot (-\sin 2x) \]
Step 2: Calculation / Simplification}
For \(0 \leq x \leq \pi/2\): \(\sin^{-1}(\sin x) = x\), \(\cos^{-1}(\cos x) = x\)
\(\frac{dy}{dx} = x \cdot \sin 2x - x \cdot \sin 2x = 0\)
\(\therefore y\) is constant (parallel to \(x\)-axis).
Step 3: Final Answer
\[ \text{Straight line parallel to the } x\text{-axis} \]
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