A) \[\frac{1}{n}\]
B) \[\frac{1}{(n-1)}\]
C) \[\frac{n}{(n-1)}\]
D) \[\frac{1}{(n-2)}\]
Correct Answer: B
Solution :
[b] Let \[{{I}_{n}}=\int\limits_{0}^{\pi /4}{{{\tan }^{n}}xdx}\] Consider, \[{{I}_{n}}+{{I}_{n-2}}=\int\limits_{0}^{\pi /4}{{{\tan }^{n}}xdx+\int\limits_{0}^{\pi /4}{{{\tan }^{n-2}}xdx}}\] \[=\int\limits_{0}^{\pi /4}{{{\tan }^{n-2}}x({{\tan }^{2}}x+1)dx}\] \[=\int\limits_{0}^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n-2}}xdx}\] Put \[\tan x=t\] \[{{\sec }^{2}}xdx=dt\] when \[x=0\] then \[t=0\] and when \[x=\frac{\pi }{4},t=1\] \[\therefore {{I}_{n}}+{{I}_{n-2}}=\int\limits_{0}^{1}{{{t}^{n-2}}dt}\] \[=\left. \frac{{{t}^{n-2+1}}}{n-2+1} \right|_{0}^{1}\left. =\frac{{{t}^{n-1}}}{n-1} \right|_{0}^{1}=\frac{1}{n-1}[1-0]=\frac{1}{n-1}\]
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