Calculate $\lim\limits_{x\rightarrow 0^+} \int\limits_0^1 \ln(1+\sin(tx))dt$ 
Calculate $$\lim_{x\rightarrow 0^+}\int_0^1\ln(1+\sin(tx))dt$$

My try:
$$\lim_{x\rightarrow 0^+} \int_0^1 \ln (1+ \sin (tx)) dt=\lim_{x\rightarrow 0^+} ([t \ln (1+\sin (tx))]^1_0 - \int_0^1 \frac{t \cos (tx) x}{1+ \sin (tx)} dt)$$
Then I want to use: $$u=1+\sin (tx), du=\cos (tx) x$$
But then I have: $$\lim_{x\rightarrow 0^+}([t \ln (1+\sin (tx))]^1_0 - \int_1^{1+\sin x} \frac{\arcsin(u-1)}{ux} du)$$
So I think my idea about $u$ is not helpfull and I need other idea.Can you help me?

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 
$$
0 \le  \int_0^1 \ln (1+ \sin (tx)) dt \le \frac x2 \, .
$$
