If [b_n = _0^ 2 ^2nx xdx,,n N ], then

Question

If $${b_n} = \int_0^{{\pi  \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N} $$, then

Accepted Answer

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

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

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

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.

Solutions · 1