If ${\int }_0^{\mathrm{\infty }} x^{2n+1}\cdot e^{-x^2}dx=360$ then the value of $n$ is

$I={\int }_0^{\mathrm{\infty }} {\left(x^2\right)}^n\cdot x{e}^{-x^2}dx$

Put  $x^2=t$ and $xdx=dt/2$ 

$\begin{array}{l}\therefore I=\frac{1}{2}{\int }_0^{\mathrm{\infty }} t^n{e}^{-t}dt\\ \left.=\frac{1}{2}{\left[-t^n{e}^{-t}\right]}_0^{\mathrm{\infty }}+n{\int }_0^{\mathrm{\infty }} t^{n-1}{e}^{-t}dt\right]\\ =\frac{1}{2}\left[0+n{\int }_0^{\mathrm{\infty }} t^{n-1}{e}^{-t}dt\right]\\ =\frac{n!}{2}=360\\ \Rightarrow n=6\end{array}$