`int_0^oox^2/(1+x^6)^ndx`

Explanation

`I=int_0^oox^2/(1+x^6)^ndx`
Let `x^3=tantheta`
`rArr3x^2dx=sec^2thetad theta`
`x=0rArrtheta=0` ,br> `x=oorArrtheta=pi//2`
`therefore I=1/3int_0^(pi/2)(sec^2thetad theta)/(sec^2theta)^n`
=`1/3int_0^(pi/2)cos^(2n-2)thetad theta`
=`1/3(2n-3)/(2n-2)cdot(2n-5)/(2n-4)...1/2cdotpi/2`