$${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $$

$$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $$

$$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$$

So, $${b_3} - {b_2}$$, $${b_4} - {b_3}$$, $${b_5} - {b_4}$$ are in H.P.

$$ \Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$$ are in A.P. with common difference $$-$$2.