Maple is not a human. A student would make the change of variables e.g. x = t^(3/4), and then compute directly.
But Maple uses sometimes lookup tables (with patterns) such as int(x^a*(1+A*x^b)^c, x), in terms of special functions.
The conversion from these special functions to elementary ones is not always easy, or even possible.

BTW, in Maple 2020 the integral is computed directly.

resultant makes sense only with respect to a single variable. So, resultant(f, g, x) eliminates (roughly) x for the system {f=0, g=0}.
If you want to eliminate both x and y, use eliminate or apply resultant twice.
Actually, you should say what exaclty you want to achieve.

int(simplify(v), x);

works.

There is a bug in Maple 2020 which does not integrate the obvious integral

int(sqrt(-x*(x - 20))*sqrt(-1/(x*(x - 20))), x = 0 .. h) assuming h>=10,h<=20;

( should be of course h) .

Workaround:

f := sqrt(-x^2+20*x):
F := simplify(f*sqrt(1+diff(f,x)^2)) assuming x>0, x<20:
solve(2*Pi*int(F, x=0..h) = 1005);

          201/(4*Pi)

It is easy to see that if there was a solution, then 
s  :=  a^4 + 4*a^3*k + 2*a^2*k^2 + 4*a*k^3 + k^4
would be an exact square for a,k positive integers (actually 0 < k < a).

But s is never a square for 0 < k < a. This can be proved mathematically, or checked by Maple for a <= N = 5000 say:

N:=5000;
seq(seq(`if`(issqr(a^4 + 4*a^3*k + 2*a^2*k^2 + 4*a*k^3 + k^4), [a,k],NULL), k=1..a-1),a=1..N);

You define u(x,t) := ...,  but u is already defined as a table u[i]:=...
This generates a total mess. Use another "variable", e.g. U(x,t) instead.

How can you believe that the expressions could be equal? Not only the exponents of sin(...) are different (2/n  versus 2/(n-1)) but the constants are not the same, one of them is piecewise, ...

C:=(m::nonnegint,n::integer) -> 
  coeff(coeff(product((1-q^i)*(1-q^i/z)*(1-q^(i-1)*z), i=1..m+1),q,m),z,n):

C(6,2),  C(6,4); 

      0,  1

(k is an arbitrary integer)

restart;
f:=(x,c,d)->(x-c)*(x+c)*(x-d)*(x+d):
g:=(x,c,d)->(x-c)*(x+c-1)*(x-d)*(x+d-1):
N:=10:
for c to N do  for d from c+1 to N do
  for x1 to c-1 do
    for F in [f,g] do
      if nops(factor(F(x,c,d)-F(x1,c,d)))=4 then lprint(F(x+k,c,d),m=F(x1,c,d)) fi;
    od
  od
od od:

(x+k-4)*(x+k+3)*(x+k-5)*(x+k+4), m = 180
(x+k-4)*(x+k+3)*(x+k-6)*(x+k+5), m = 360
(x+k-4)*(x+k+4)*(x+k-7)*(x+k+7), m = 720
(x+k-5)*(x+k+4)*(x+k-7)*(x+k+6), m = 504
(x+k-5)*(x+k+4)*(x+k-9)*(x+k+8), m = 1260
(x+k-5)*(x+k+4)*(x+k-10)*(x+k+9), m = 1800
(x+k-5)*(x+k+5)*(x+k-10)*(x+k+10), m = 2016
(x+k-6)*(x+k+5)*(x+k-7)*(x+k+6), m = 1260
(x+k-6)*(x+k+6)*(x+k-7)*(x+k+7), m = 1440
(x+k-6)*(x+k+5)*(x+k-9)*(x+k+8), m = 1080
(x+k-7)*(x+k+7)*(x+k-9)*(x+k+9), m = 2880
(x+k-7)*(x+k+6)*(x+k-10)*(x+k+9), m = 3780
(x+k-8)*(x+k+8)*(x+k-9)*(x+k+9), m = 5040
(x+k-9)*(x+k+8)*(x+k-10)*(x+k+9), m = 5544
(x+k-9)*(x+k+8)*(x+k-10)*(x+k+9), m = 2520
 

1. In Maple 2020 it can be done because a (statement)  is an expression

restart;
x:=10: str:="A":
str:=cat(str,  `if`(x=10,  [(x:=11)," it was 10"][2], [(x:=8)," it was not 10"][2]));  x;

2. `if`  is alias for ifelse  and a link appears in the help page of if. Or, use ?ifelse 

p:=randpoly([x,y,z]);
`%+`(sort([op(p)], key=abs@coeffs)[]);

Use value(%)  to  go back.

This isthe well known  Pell's equation.

S:=isolve(61*x^2+1=y^2):
for _Z1 from 0 to 6 do
lprint(eval([x,y],rationalize(S[4]))[])
od;

0, 1
226153980, 1766319049
798920165762330040, 6239765965720528801
2822295814832482312327709940, 22042834973108102061352541449
9970149719303180503641083029374964080, 77869358613928486808166555366140995201
35220930741174421456911021812718768924061809900, 275084262906388245923976756042747916825335226249
124422801783292138491822391332416163557158135530198606120, 971773147303355325052564141449134520779147876502526039201
 

It is easy to obtain such equalities. Here is a simple example. It can be extended to a generic one.
Why do you consider them so important?

g:=proc(z)
   local m:=subs(sqrt(3)=0,z),
         n:=z-m;
   m^2+n^2-1+ 2*m*n-sqrt(3)
end: 
F:=c->sqrt(1+sqrt(3)+c):
G:= n -> (F@@n)(g@@n)(1+sqrt(3))=1+sqrt(3):
G(3);G(4);G(5);

You do not have a value (true or false) for print_table
Note also that a better  header for your proc is:

proc(f::procedure, a::realcons, b::realcons, N::posint, print_table::truefalse)
 

You must use parallel substitutions in ex:
ex := simplify(subs([ X = cos(theta)*X-sin(theta)*Y, Y = sin(theta)*X+cos(theta)*Y ], eq)); 

(You have defined the procedure f. It is then easier to use f(X-9/7, Y+8/7)  and similar for ex, instead of subs) .