Question

how do i integrate :

integral from 0 to 1 of pi[4tan(pi/3*y)]^2

Answer 1

16pi tan^2(y pi/3)= 16 pi (sec^2(y pi / 3)-1) then you can go from there, i think amistre and i did this one for you earlier

Answer 2

u = pi / 3 y
3 / pi du= dy
INT 3 / pi (16pi)sec^2(u)-1 dy
48(tan(u)-y)+C
since u=pi/3y
48((tan(pi / 3 y)-y)+C

Answer 3

then from [0,1]
since tan(0)=0 and y=0 just use the top bound to evaluate the integral
48(sqrt(3)-1)