For n geq 2,let I_n = int_0^{pi / 4} tan^n x | vedclass

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(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$ for $n \geq 2$. $F_n = I_n + I_{n-2} = \int_0^{\pi / 4} (	an^n x + 	an^{n-2} x) \, dx$. $F_n = \in

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$\frac{1}{n(n-1)}$

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(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$ for $n \geq 2$. $F_n = I_n + I_{n-2} = \int_0^{\pi / 4} (	an^n x + 	an^{n-2} x) \, dx$. $F_n = \int_0^{\pi / 4} 	an^{n-2} x (	an^2 x + 1) \, dx = \int_0^{\pi / 4} 	an^{n-2} x \sec^2 x \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. When $x = 0, t = 0$ and when $x = \pi / 4, t = 1$. $F_n = \int_0^1 t^{n-2} \, dt = \left[ \frac{t^{n-1}}{n-1} ight]_0^1 = \frac{1}{n-1}$. Similarly,$F_{n+1} = \frac{1}{(n+1)-1} = \frac{1}{n}$. Therefore,$F_n - F_{n+1} = \frac{1}{n-1} - \frac{1}{n} = \frac{n - (n-1)}{n(n-1)} = \frac{1}{n(n-1)}$.

For $n \geq 2$,let $I_n = \int_0^{\pi / 4} \tan^n x \, dx$ and $F_n = I_n + I_{n-2}$. Then,$F_n - F_{n+1} =$

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