If $\displaystyle I_n = \int_{0}^{n\pi} \frac{\sin x}{1+x}dx$ then I was asked to find $I_1,I_2,I_3,I_4$ in increasing order of magnitude for $n=1,2,3,4$.
My approach : Start with $\displaystyle I_{n+1}-I_{n}=\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{1+x}dx$.
Now since $-1\le \sin x \le 1$ , So $\frac{-1}{1+x}\le \frac{\sin x}{1+x}\le \frac{1}{1+x}$.
So integrating each side with proper limits we obtain $$\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{1+x}\le \ln(1+\frac{1}{n})<0.$$
So, $I_n$ is a decreasing integral since $I_{n+1}<I_n$ and so $I_4<I_3<I_2<I_1$.
Is this approach ok ?