If int_0^{pi / 2} tan ^{14}left(frac{x}{2}rig | vedclass

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(C) Let $I = \int_0^{\pi / 2} 	an^{14}(\frac{x}{2}) dx$. Let $t = \frac{x}{2}$, then $dx = 2dt$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\frac{\

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$\frac{(-1)^{n+1}}{2 n-1}$

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(C) Let $I = \int_0^{\pi / 2} 	an^{14}(\frac{x}{2}) dx$. Let $t = \frac{x}{2}$, then $dx = 2dt$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\frac{\pi}{4}$.$I = 2 \int_0^{\pi/4} 	an^{14}(t) dt$.Using the reduction formula $I_n = \int 	an^n(t) dt = \frac{	an^{n-1}(t)}{n-1} - I_{n-2}$, we have $\int_0^{\pi/4} 	an^n(t) dt = \frac{1}{n-1} - \int_0^{\pi/4} 	an^{n-2}(t) dt$.Applying this repeatedly for $n=14, 12, \dots, 2$:$I_{14} = \frac{1}{13} - I_{12} = \frac{1}{13} - (\frac{1}{11} - I_{10}) = \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} + \frac{1}{5} - \frac{1}{3} + \frac{1}{1} - \int_0^{\pi/4} 1 dt$.Since $\int_0^{\pi/4} 1 dt = \frac{\pi}{4}$, we get $I = 2 [ \sum_{k=1}^7 \frac{(-1)^{k+1}}{2k-1} - \frac{\pi}{4} ]$.Comparing this with the given expression $2 [ \sum_{n=1}^7 f(n) - \frac{\pi}{4} ]$, we find $f(n) = \frac{(-1)^{n+1}}{2n-1}$.

If $\int_0^{\pi / 2} \tan ^{14}\left(\frac{x}{2}\right) d x=2\left[\sum_{n=1}^7 f(n)-\frac{\pi}{4}\right]$, then $f(n)=$

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