Calculate $\lim\limits_{x\rightarrow 0^+} \int\limits_0^1 \ln(1+\sin(tx))dt$

It is not necessary to compute the integral explicitly. The following estimate is sufficient to determine the limit:

If $0 \le x \le \pi$ then $\sin(tx) \ge 0$ for all $t \in [0, 1]$, so that $$ 0 \le \ln (1+ \sin (tx)) \le \sin(tx) \le tx $$ and therefore $$ 0 \le \int_0^1 \ln (1+ \sin (tx)) dt \le \frac x2 \, . $$