The antiderivative cannot be calculated using u-subs, trig-subs, integration by parts, or other similar methods of integration, so I used a series representation to integrate.
#int (20sin(t^2/35))dt#
#=20int(sin(t^2/35))dt#
A u-substitution would not work for this problem because the #u# would not cancel. So, another method is to Find the MacLaurin series representation of #sin(t^2/35)#, and integrate the summation. (The coefficient #20# won't change)
Remember: #color(blue)(sin u =sum_(n=0)^(oo) frac{(-1)^n (u)^(2n+1)}{(2n+1)!})#
Substitute #u=t^2/35#
#sin(t^2/35)=sum_(n=0)^(oo) frac{(-1)^n (t^2/35)^(2n+1)}{(2n+1)!}#
Expand the terms so that there is a single term with only #t# as the base of the exponent:
#sin(t^2/35)= sum_(n=0)^(oo) frac{(-1)^n(t^2)^(2n+1)}{(35)^(2n+1)(2n+1)!}#
#sin(t^2/35)=sum_(n=0)^(oo) frac{(-1)^n(t)^(4n+2)}{(35)^(2n+1)(2n+1)!}#
Now find the integral of the original problem:
#20int(sin(t^2/35))dt=20int (sum_(n=0)^(oo) frac{(-1)^n(t)^(4n+2)}{(35)^(2n+1)(2n+1)!}) dt#
Use the power rule for integrals (which is the opposite of the power rule for derivatives):
#=20(sum_(n=0)^(oo) frac{(-1)^n (t)^(4n+3)}{(35)^(2n+1)(2n+1)!(4n+3)} )+C#
