Calculus inequality (easy) I wanna prove that 
$$\forall x>1,\quad\int_{1}^{x} \frac{\sin(t)}{t} dt - x +1 < 0.$$
Is it true that I can rewrite the inequality as 
$$\int_{1}^{x} \left(\frac{\sin(t)}{t}-t \right)dt < 0,$$ 
and since $\frac{\sin(t)}{t}<t$ for $t>1$ (trivial), the inequality holds? 

Or let $f(x)=\int_{1}^{x} \frac{\sin(t)}{t} dt - x +1 $. Then $f(1)=0$ and $f'(x)=\frac {\sin(x)}{x}-1$ and it's is known that $\sin(x)< x$ for $x>0$. Thus $f'(x)< 0$ and $f$ is strictly decreasing. So $f(x)<f(1)=0$ for $x> 1$.
