The previously stated value of 1.06484 is not correct. The integral from 0 to 1 has to be less that 1, because sin t/t never exceeds 1.
One way to do the integration would be to use the integration capability that some hand calculators have. Another would be to perform a "numerical" integration using Simpson's rule. A third way, which I show below, is to represent sin t by the first few terms of an infinite series.
sin t = t - t^3/6 + t^5/120 - t^7/5040...
(sin t)/t = 1 - t^2/6 + t^4/120 -t^6/5040
Integral from 0 to 1 = 1 - 1/(3*6) + 1/(120*5) - 1/(5040*7)..
= 1 - 0.05556 + 0.00139 - 0.00003 = 0.94586
Using Simpson's rule and numerical values of sin t/t at 0, 0.25, 0.5, 0.75 and 1 gives a value of 0.94609.
Higher accuracy can always be obtained with either method by using more terms, but to three significant figures the answer is 0.946.
Thanks for this opportunity to help you. If my answer was helpful, please click the Accept button. If not, please click Reply.