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$I = \vert\int_2^5 \frac{\sin{x}\;dx}{1+x^2} \vert$ , then

$\\(a)\;\; I \ge \frac{1}{4}$

$\\(b)\;\;\;I$ lies in the interval $(\frac{1}{4},\;\frac{1}{5})$

$\\(c)\;\;\;I$ lies in the interval $(\frac{1}{5},\;\frac{1}{6})$

$\\(d)\;\; I \le \frac{3}{10}$

I started applying integration by parts, but realized it may not be the correct approach. How to proceed on evaluating this integral

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    From the nature of the possible answers I expect you are meant to estimate the integral numerically, not try to get an exact answer for it. – 2017-01-22
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    Your title is wrong. You are not evaluating the integral, you are trying to answer a question about it. – 2017-01-22
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    sure... modified to approximate the integral – 2017-01-23

3 Answers 3

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Assuming it is a test question and only one option is correct, the only correct option is $d)$.
The integrand function is pretty small in absolute value and has a sign change at $\pi\in[2,5]$, hence $a)$ is wrong for sure. If $b)$ or $c)$ were correct, $d)$ would be correct too. Numerically the integral is $\approx \frac{8}{385}$.

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To show that d is the answer we just have to show that the integral is less than $\frac 16$. We can do that by breaking it into two integrals and bounding each $$\begin {align}\int_2^5 \frac{\sin{x}\;dx}{1+x^2}&=\int_2^\pi \frac{\sin{x}\;dx}{1+x^2}+\int_\pi^5 \frac{\sin{x}\;dx}{1+x^2}\\ &\lt\int_2^\pi \frac{1\;dx}{1+x^2}+0\\ &\lt \arctan \pi - \arctan 2 \\ &\lt 0.156 \end {align}$$

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$I = \vert\int_2^5 \frac{\sin{x}\;dx}{1+x^2} \vert \le \int_2^5 \frac{\vert\sin{x}\vert\;dx}{1+x^2} \le \int_2^5 \frac{1\;dx}{1+x^2} \le \int_2^5 \frac{1\;dx}{x^2}$

Now, $\int_2^5 \frac{1\;dx}{x^2} = -\frac{1}{n}|_2^5 = \frac{1}{2}-\frac{1}{5} = \frac{3}{10}$

$I \le \frac{3}{10}$