prooove

\int_{0}^{n\pi }{\left|\frac{sinx}{x} \right|}dx = \int_{0}^{\pi }{\left|\frac{sinx}{x} \right|}dx + \int_{\pi }^{2\pi }{\left|\frac{sinx}{x} \right|}dx...........\int_{(n-1)\pi }^{n\pi }{\left|\frac{sinx}{x} \right|}dx
\int_{0}^{\pi }{\frac{sinx}{x}}dx \dx + \int_{0}^{\pi }{\left|\frac{sin(t+\pi )}{t+\pi } \right|}dt + \int_{0}^{\pi }{\left|\frac{sin(u+2\pi )}{u+2\pi } \right|}du...............
=\sum_{r=1}^{n}{\int_{0}^{\pi }{\frac{sinx}{x+(r-1)\pi } }}dx > \sum_{r=1}^{n}{\int_{0}^{\pi }{\frac{sinx}{\pi +(r-1)\pi } }}dx

=\sum_{r=1}^{n}{\int_{0}^{\pi }{\frac{sinx}{\pi r } }}dx=\sum_{r=1}^{n}{\frac{2}{\pi r}}=\frac{2}{\pi }\left(1+\frac{1}{2}+\frac{1}{3}.......\frac{1}{n} \right)

P.S this is a fiitjee aits q

i already did it half way..but plz explain 3rd step

see....
here ∩ is \pi

in lhs since x is varying from 0→∩

so for every value of x from 0→∩

sin xx>sin x∩

so sin xx+(r-1)∩>sin x∩+(r-1)∩

therefore

∫(0→∩)sin xx+(r-1)∩>∫(0→∩)sin x∩+(r-1)∩

hence

\sum_{r=1}^{n}{\int_{0}^{\pi }{\frac{sinx}{x+(r-1)\pi } }}dx > \sum_{r=1}^{n}{\int_{0}^{\pi }{\frac{sinx}{\pi +(r-1)\pi } }}dx

i hope i m correct with my explanation :P