Calculus inequality (easy)

Question

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?

Answer 1

Since $\sin(t)\leq 1$, $$\int_{1}^{x}\frac{\sin t}{t}\,dt < \int_{1}^{x}\frac{dt}{t}=\log x,$$ hence you just need to prove that for any $x>1$: $$ \log x < x-1 $$ or that for any $y>0$ we have: $$ \log(y+1) < y $$ that is equivalent to: $$ e^y > 1+y,$$ well-known.

Answer 2

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 $x> 1$.