If I_n = int_0^{pi / 4} tan^n x , dx, then I_ | vedclass

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(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider the sum $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \, dx + \int_0^{\pi / 4} 	an^{r+2} x \

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harmonic progression

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(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider the sum $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \, dx + \int_0^{\pi / 4} 	an^{r+2} x \, dx$. $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x (1 + 	an^2 x) \, dx$. Since $1 + 	an^2 x = \sec^2 x$, we have $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \sec^2 x \, dx$. Let $t = 	an x$, then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$. Thus, $I_r + I_{r+2} = \int_0^1 t^r \, dt = \left[ \frac{t^{r+1}}{r+1} ight]_0^1 = \frac{1}{r+1}$. For $r=2$, $I_2 + I_4 = \frac{1}{3}$. For $r=3$, $I_3 + I_5 = \frac{1}{4}$. For $r=4$, $I_4 + I_6 = \frac{1}{5}$. The sequence is $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$, which are in harmonic progression $(HP)$ because their reciprocals $3, 4, 5, \ldots$ are in arithmetic progression.

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$, then $I_2+I_4, I_3+I_5, I_4+I_6, \ldots$ are in

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