How is this integration possible?

First I will show how to derive the definite integral form in question: Consider the integral $I=\displaystyle \int_{0}^{a} e^{x^{2}}$ , where a is an arbitrary number. Now the double integral $\displaystyle \int_{0}^{a} \int_{0}^{a} e^{x^{2}+y^{2}} dxdy=\displaystyle \int_{0}^{a} e^{x^{2}} dx \int_{0}^{a} e^{y^{2}}dy$ , because each iteration evaluates to a constant number which does not affect the final value of the integral other than scaling. We have defined $\displaystyle \int_{0}^{a} e^{x^{2}}=I$ , then $\displaystyle \int_{0}^{a} \int_{0}^{a} e^{x^{2}+y^{2}} dxdy=\displaystyle \int_{0}^{a} e^{x^{2}} dx \int_{0}^{a} e^{y^{2}}dy=I^{2}$

Note that this integral is done over the square region in the first quadrant with side length $a$ .

Now for $a=1$ we have , $I^{2}=\displaystyle \int_{0}^{1} \int_{0}^{1} e^{x^{2}+y^{2}} dxdy = \left(\int_{0}^{1} e^{x^{2}} dx\right)^{2}=2 \displaystyle \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sec \theta} e^{r^{2}} rdrd\theta$ .

From $\theta =$ 0 to $\frac{\pi}{4}$ the projection of $r$ variable to x-axis ,which is the distance from origin to the side, will be always 1, which gives $rcos\theta =1$ and $r=\sec \theta$ . The whole integral is multiplied by 2 since we need to cover the whole square region ,from $\theta=0 to \frac{\pi}{2}$ , and from symmetry arguments $r$ takes the same continuous values but decreasing from $\theta =\frac{pi}{4}$ to $\frac{\pi}{2}$ .

Now , $2 \displaystyle \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sec \theta} e^{r^{2}} rdrd\theta=2 \displaystyle \int_{0}^{\frac{\pi}{4}} \left( \frac{e^{\sec^{2} \theta} -1}{2}\right) d\theta =\displaystyle \int_{0}^{\frac{\pi}{4}} e^{\sec^{2} \theta}d\theta -d\theta$

$\displaystyle \int_{0}^{\frac{\pi}{4}} e^{\sec^{2} \theta}d\theta -d\theta=\left(\int_{0}^{1} e^{x^{2}} dx\right)^{2}$

$\left(\displaystyle \int_{0}^{\frac{\pi}{4}} e^{\sec^{2} \theta}d\theta\right) -\frac{\pi}{4}=\left(\displaystyle \int_{0}^{1} e^{x^{2}} dx\right)^{2}$

$\left(\displaystyle \int_{0}^{\frac{\pi}{4}} e^{\sec^{2} \theta}d\theta\right) =\left(\displaystyle \int_{0}^{1} e^{x^{2}} dx\right)^{2}+\frac{\pi}{4}$

From $1+\tan^{2} \theta =\sec^{2} \theta$

$\displaystyle \int_{0}^{\frac{\pi}{4}} e^{sec^{2} \theta}d\theta =\displaystyle \int_{0}^{\frac{\pi}{4}} e^{1+tan^{2} \theta}d\theta=e\displaystyle \int_{0}^{\frac{\pi}{4}} e^{tan^{2} \theta}d\theta$

$e\displaystyle \int_{0}^{\frac{\pi}{4}} e^{tan^{2} \theta}d\theta=\left(\displaystyle \int_{0}^{1} e^{x^{2}} dx\right)^{2}+\frac{\pi}{4}$

$\displaystyle \int_{0}^{\frac{\pi}{4}} e^{tan^{2} \theta}d\theta=\frac{\left(\displaystyle \int_{0}^{1} e^{x^{2}} dx\right)^{2}+\frac{\pi}{4}}{e}$

$\displaystyle \int_{0}^{\frac{\pi}{4}} e^{tan^{2} \theta}d\theta=\left(\displaystyle \int_{0}^{1} e^{x^{2}-\frac{1}{2}} dx\right)^{2}+\frac{\pi}{4e}$

Adding up the constants as shown in the question will give 10.5