Given function : ∫ 4a²x⁶e^−x2 \ \

\ \

First derive general solution for

Let u = x^2n-1

u =(2n-2) x^2n-2

Let dv=xe^−x2

v = -½ e^−x2

Changing to polar coordinates :

If r² = u then rdr = du/2 \ \

Here angle gives π degrees.So

From hypothesis rule f is at 0 and n are derived

The general solution

Given problem is

Here n=3

The solution is ∫ 4a²x⁶e^−x2 dx = (15/16)√π \ \