If b_n = _0^/2(2nx)/( x),dx, nN, then:

Question

If $b_n = \displaystyle\int_0^{\pi/2}\frac{\cos2nx}{\sin x}\,dx$, $n\in\mathbb{N}$, then:

Accepted Answer

$$b_n-b_{n-1} = \int_0^{\pi/2}\frac{\cos2nx-\cos2(n-1)x}{\sin x}\,dx = \int_0^{\pi/2}\frac{-\sin(2n-1)x\sin x}{\sin x}\,dx = -\int_0^{\pi/2}\sin(2n-1)x\,dx$$

$$= \left[\frac{\cos(2n-1)x}{2n-1}\right]_0^{\pi/2} = -\frac1{2n-1}$$

So $b_3-b_2=-\dfrac15,\;b_4-b_3=-\dfrac17,\;b_5-b_4=-\dfrac19$ --- an H.P. --- hence their reciprocals $-5,-7,-9$ are in A.P. with common difference $-2$.

Solutions · 1