If I_{n}=int_{0}^{pi / 4} tan ^{n} x d x,wher | vedclass

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(A) Given integral is $I_{n} = \int_{0}^{\pi / 4} 	an ^{n} x d x$. We can write $	an^{n} x$ as $	an^{n-2} x \cdot 	an^{2} x$. Using the identity $

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$\frac{1}{9}$

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(A) Given integral is $I_{n} = \int_{0}^{\pi / 4} 	an ^{n} x d x$. We can write $	an^{n} x$ as $	an^{n-2} x \cdot 	an^{2} x$. Using the identity $	an^{2} x = \sec^{2} x - 1$,we have: $I_{n} = \int_{0}^{\pi / 4} 	an^{n-2} x (\sec^{2} x - 1) d x$. $I_{n} = \int_{0}^{\pi / 4} 	an^{n-2} x \sec^{2} x d x - \int_{0}^{\pi / 4} 	an^{n-2} x d x$. Let $t = 	an x$,then $dt = \sec^{2} x dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$. $I_{n} = \int_{0}^{1} t^{n-2} dt - I_{n-2}$. $I_{n} = \left[ \frac{t^{n-1}}{n-1} ight]_{0}^{1} - I_{n-2}$. $I_{n} = \frac{1}{n-1} - I_{n-2}$. Therefore,$I_{n} + I_{n-2} = \frac{1}{n-1}$. For $n=10$,we get $I_{10} + I_{8} = \frac{1}{10-1} = \frac{1}{9}$.

If $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$,where $n$ is a positive integer,then $I_{10}+I_{8}$ is

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