\(I=\)\(∫_0^{\frac \pi2}cos^2x\ dx\) ...(1)
⇒ \(I\) =\(∫_0^{\frac \pi2}cos^2(\frac \pi2-x)\ dx\) \((∫_0^aƒ(x)dx = ƒ(a-x)dx)\)
⇒\(I\) = \(∫_0^{\frac \pi2}sin^2x\ dx\) ...(2)
Adding (1) and (2), we obtain
\(2I\) =\(∫_0^{\frac \pi2}(sin ^2x+cos^2x)\ dx\)
⇒\(2I\) = \(∫_0^{\frac \pi2}1\ dx\)
⇒\(2I\) = \([x]_0^{\frac \pi2}\)
⇒\(2I\) = \(\frac \pi2\)
⇒\(I\) = \(\frac \pi4\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
