For the function $f(x)=\int\limits_0^x \frac{\sin t}{t} d t$, where $x>0$. At $x=n \pi, f(x)$ attains

maximum or minimum according as $n$ is odd or even respectively

We have,

$f(x) =\int\limits_0^x \frac{\sin t}{t} d t$

$\Rightarrow f'(x) =\frac{\sin x}{x}$ and $f''(x)=\frac{x \cos x-\sin x}{x^2}$

For maximum/minimum, we must have

$f'(x)=0$

$\Rightarrow \frac{\sin x}{x} =0 \Rightarrow \sin x=0 \Rightarrow x=n \pi, n \in N$      [∵ x > 0]

At $x=n \pi$, we have

$f''(x)=\frac{n \pi \cos n \pi-\sin n \pi}{n^2 \pi^2}=\frac{(-1)^n}{n \pi}$

$\Rightarrow f''(x) < 0$, if n is an odd natural number 

and, f''(x) > 0, if n is an even natural number.

Hence, f(x) attains local maximum or minimum at $x=n \pi$ according as n is an odd natural number or an even natural number.