The Correct Answer is:\(I = \frac{\pi}{4}\)
Let \(I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}xdx}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x} ...(1)\)
⇒\(I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}(\frac{\pi}{2} - x)}{sin^{\frac{3}{2}}(\frac{\pi}{2} - x)+cos^{\frac{3}{2}}(\frac{\pi}{2} - x)} dx \,\,\,\,\,\,\, (\int^a_0 f(x)dx= \int^a_0 f(a-x)dx)\)
⇒\(I = \int^{\frac{π}{2}}_0 \frac{cos^{\frac{3}{2}}x}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}dx...(2)\)
Adding(1)and(2),we obtain
\(2I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}dx\)
⇒\(2I = \int^{\frac{π}{2}}_0 1.dx\)
⇒\(2I = [x]^\frac{\pi}{2}_0\)
⇒\(2I = \frac{\pi}{2}\)
⇒\(I = \frac{\pi}{4}\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
