We now apply the trigonometric identity $1ta{n}^2()=se{c}^2()$ and simplify the result.

$\frac{2sin()(1ta{n}^2())}{se{c}^2()}=\frac{()sin()se{c}^2()}{se{c}^2()}$

$=()sin()$

This allows us to rewrite the given integral as follows.

${\displaystyle \underset{0}{\overset{4\frac{}{3}}{}}}\frac{2sin()(1ta{n}^2())}{se{c}^2()}d={\displaystyle \underset{0}{\overset{4\frac{}{3}}{}}}()sin()d$