Solutions:

Let $\quad 1=\int_{0}^{\infty} 3^{-4 x^{2}} d x$

put $\quad 3^{-4 x^{2}}=e^{-t}$

taking log on both sides,

$4 x^{2} \log 3=t$

$x^{2}=\frac{t}{4 \log 3} \quad=\gt\quad x=\frac{\sqrt{t}}{2 \sqrt{\log 3}}$

diff w.r.t x,

$d x=\frac{t^{-1 / 2}}{4 \sqrt{\log 3}} d t$

$\therefore \mathrm{I}=\int_{0}^{\infty} \frac{e^{-t}}{4 \sqrt{\log 3}} t^{-1 / 2}$

$\therefore \mathrm{I}=\frac{1}{4 \sqrt{\log 3}} \int_{0}^{\infty} e^{-t} \cdot t^{-1 / 2} \mathrm{dt}$

$\therefore \mathrm{I}=\frac{\sqrt{\pi}}{4 \sqrt{\log 3}}$ $\ldots \ldots \ldots\left\{\int_{0}^{\infty} e^{-t} \cdot t^{-1 / 2} d t=\sqrt{\pi}\right\}$