Let I=\(∫^\frac{π}{2}_0\frac{sinx-cosx}{1+sinxcosxd}x........(1)\)
\(⇒I=∫^\frac{π}{2}_0\frac{sin(\frac{π}{2}-x)-cos(\frac{π}{2}-x)}{1+sin(\frac{π}{2}-x)cos(\frac{π}{2}-x)dx (∫a0ƒ(x)dx=∫a0ƒx)}dx)\)
\(⇒I=∫_0^{π}{2}\frac{cosx-sinx}{1+sinxcosx}dx...(2)\)
\(Adding(1)and(2),we obtain\)
\(2I=∫_0^\frac{π}{2}\frac{0}{1+sinxcosx}dx\)
\(⇒I=0\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
