Prove that $$\int_0^{\pi/2} e^{-\tan^2(x)}\,dx=\frac{\pi e}{2}\big(1-\operatorname{erf}(1)\big)$$
My Attempt
Let $u=\tan(x)\implies du=\sec^2(x)\,dx=1+u^2\,dx$
$$I=\int_0^{\pi/2} e^{-\tan^2(x)}\,dx=\int_0^\infty \frac{e^{-u^2}}{1+u^2}\,du$$
How do you proceed from here? I was thinking of integrating by parts but it doesn't seem to work. How do you solve this integral? Thank you for your time
Note that, $$\int_{0}^{\infty } e^{-(1+t^{2})y} dy=\frac{1}{(1+t^{2})}\\ $$ $$\int_{0}^{\infty }e^{-t^{2}a}dt=\frac{^{\sqrt{\pi}}}{2\sqrt{a}}$$ $$\int_{0}^{z}e^{-t^{2}}dt=\frac{^{\sqrt{\pi}}}{2}erf(z)$$ Therefore, the solution is $$ \int_{0}^{\infty }\frac{e^{-t^{2}}}{(1+t^{{2}})}dt=\int_{0}^{\infty }\int_{0}^{\infty } e^{-(1+t^{2})y}e^{-t^{2}}dtdy$$$$\int_{0}^{\infty }\int_{0}^{\infty }e^{-t^{2}(1+y)}e^{-y}dtdy=\int_{0}^{\infty }\frac{^{\sqrt{\pi}}e^{-y}}{2\sqrt{1+y}}dy $$ $$\sqrt{1+y}=x$$$\sqrt{\pi}\int_{1}^{\infty }e^{-x^{2}+1}dx=\frac{\pi}{2}e(1-\operatorname{erf}(1))$