${\int }_0^{\frac{\pi }{4}}{e}^{{\tan }^2\theta }{\sin }^2\theta \tan \theta d\theta =$

$I={\int }_0^{\frac{\pi }{4}}{e}^{{\tan }^2\theta }{\sin }^2\theta \tan \theta d\theta$

$={\int }_0^{\frac{\pi }{4}}{e}^{{\tan }^2\theta }\frac{{\tan }^3\theta {\sec }^2\theta }{{\left(1+{\tan }^2\theta \right)}^2}d\theta$

Let $\tan \theta =t$

$I={\int }_0^1\frac{e^{t^2}{t}^3}{{\left(1+t^2\right)}^2}dt$

Let $t^2=u$

$I={\int }_0^1\frac{e^{u}u}{2{\left(1+u\right)}^2}du=\frac{1}{2}{\int }_0^1{e}^{u}u\frac{1}{{\left(1+u\right)}^2}du$

By integrating by parts method

$I=\frac{1}{2}{\left[e^{u}u\left(\frac{-1}{1+u}\right)-{\int }_0^1\left(e^u+e^{u}u\right)\left(\frac{-1}{1+u}\right)du\right]}_0^1$

$=\frac{1}{2}{\left[e^{u}u\left(\frac{-1}{1+u}\right)+{\int }_0^1{e}^{u}du\right]}_0^1=\frac{1}{2}{\left[\frac{-e^{u}u}{1+u}+e^u\right]}_0^1$

$=\frac{1}{2}\left(\frac{-e}{2}+e-1\right)=\frac{1}{2}\left(\frac{e}{2}-1\right)$