ln(sec2x)^1/x^2
You asked:
Evaluate the expression: \(\frac{{\ln\left( \sec\left( 2 x \right) \right)}^{1}}{{x}^{2}}\)
MathBot Answer:
\[\frac{{\ln\left( \sec\left( 2 x \right) \right)}^{1}}{{x}^{2}} = \frac{\log{\left(\sec{\left(2 x \right)} \right)}}{x^{2}}\]