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

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 $$I_{n+1} and so $$I_4.

Is this approach ok ?

• It has been answered here also: math.stackexchange.com/questions/1843585/… Nov 26, 2016 at 13:12
• Ya but I thought this approach might work so I posted it. Nov 26, 2016 at 13:15
• Could have answered that question rather :) No problem. Nov 26, 2016 at 13:16
• Found this question in the ISI Test of maths book, an interesting problem so I posted it Nov 26, 2016 at 13:19

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 $$I_1>I_3>I_2>I_4.$$