If I_n = _/4^/2^nx,dx, then:

Question

If $I_n = \displaystyle\int_{\pi/4}^{\pi/2}\cot^nx\,dx$, then:

Accepted Answer

$$I_n+I_{n-2} = \int_{\pi/4}^{\pi/2}\cot^{n-2}x\csc^2x\,dx$$

With $t=\cot x$: $I_n+I_{n-2} = \displaystyle\int_0^1t^{n-2}\,dt = \dfrac1{n-1}$.

So $I_2+I_4=\dfrac13$, $I_3+I_5=\dfrac14$, $I_4+I_6=\dfrac15$, giving reciprocals $3,4,5$ --- an A.P.

Solutions · 1