\(Let \space I =\int_{0}^{\frac{\pi}{4}}(\frac{sinxcosx}{cos^{4}x+sin^{4}x})dx\)
\(⇒I=\int_{0}^{\frac{\pi}{4}}\frac{\frac{(sinxcosx)}{cos^{4}x}}{\frac{(cos^{4}x+sin^{4}x)}{cos^{4}x}}dx\)
\(⇒I=\int_{0}^{\frac{\pi}{4}}\frac{tanxsec^{2}x}{1+tan^{4}x}dx\)
\(Let \space tan^{2}x=t⇒2tanxsec^{2}xdx=dt\)
\(When x=0,t=0 \space and \space when \space x=\frac{\pi}{4},t=1\)
\(∴I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^{2}}\)
\(=\frac{1}{2}[tan^{-1}t]_{0}^{1}\)
\(=\frac{1}{2}[tan^{-1}1-tan^{-1}0]\)
\(=\frac{1}{2}[\frac{\pi}{4}]\)
\(=\frac{\pi}{8}\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]