Start from the MacLaurin series of #sinh x#:
#sinhx = sum_(n=0)^oo x^(2n+1)/((2n+1)!)#
Using the ratio test:
#lim_(n->oo) abs((a_(n+1))/(a_n)) = lim_(n->oo) abs( ((x^(2(n+1)+1)/((2(n+1)+1)!))/(x^(2n+1)/((2n+1)!)))#
#lim_(n->oo) abs((a_(n+1))/(a_n)) = lim_(n->oo) abs( x^(2n+3)/x^(2n+1)) ((2n+1)!)/((2n+3)!)#
#lim_(n->oo) abs((a_(n+1))/(a_n)) = x^2 lim_(n->oo) 1/((2n+3)(2n+2)) = 0#
we can see that the series is absolutely convergent for every #x in RR#.
Substituting #x=t^2# we have:
#sinh t^2 = sum_(n=0)^oo (t^2)^(2n+1)/((2n+1)!) = sum_(n=0)^oo t^(4n+2)/((2n+1)!) #
and as the series has radius of convergence #R=oo# we can integrate term by term and obtain a series that still has #R=oo#:
#int_0^x sinh t^2 dt = sum_(n=0)^oo int_0^x t^(4n+2)/((2n+1)!)dt #
#int_0^x sinh t^2 dt = sum_(n=0)^oo t^(4n+3)/((4n+3)(2n+1)!)#
