Now rewrite the entire expression in terms of $u$:
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- 2\sec x \csc x = \frac{2}{\sin x \cos x}
- And $2\sin x \cos x$ remains.
- Let $u = \sin x \cos x$. Since $x \in (0, \frac{\pi}{2})$, $u > 0$, and $u = \frac{1}{2} \sin 2x$, so $0 < u \leq \frac{1}{2}$.
- \frac{1}{u^2} + \frac{2}{u} + 1 + 2u
- Define the function:
- f(u) = \frac{1}{u^2} + \frac{2}{u} + 1 + 2u, \quad 0 < u \leq \frac{1}{2}
