Arrange $ I_n = \int_{0}^{n\pi} \frac{\sin x}{1+x}dx$ in increasing order of magnitude for $n=1,2,3,4$

Question

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 ?

Answer 1

Hint. We have that $$0\leq (-1)^n(I_{n+1}-I_n)=(-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{1+x}dx=\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{1+x}\,dx\leq \int_{n\pi}^{(n+1)\pi}\frac{dx}{1+x}.$$ So the increment $(I_{n+1}-I_n)$ changes sign and you should find that $$I_1>I_3>I_2>I_4.$$