A) 1/4
B) 1/5
C) 1/6
D) 1/7
Correct Answer: D
Solution :
Given, \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta .d\theta }\] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n-2}}\theta .{{\tan }^{2}}\theta .d\theta }\] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n-2}}\theta .({{\sec }^{2}}\theta -1)d\theta }\] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n-2}}\theta .{{\sec }^{2}}\theta .d\theta }-\int_{0}^{\pi /4}{{{\tan }^{n-2}}\theta .d\theta }\] \[{{I}_{n}}=\int_{0}^{1}{{{\operatorname{t}}^{n-2}}}.dt-{{I}_{n-2}}\] \[{{I}_{n}}+{{I}_{n-2}}=\left[ \frac{{{t}^{n-1}}}{n-1} \right]_{0}^{1}\] (let\[t=\tan \theta ,dt={{\sec }^{2}}\theta .d\theta \]) \[{{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-1}\] Put \[n=8\] \[{{I}_{8}}+{{I}_{6}}=\frac{1}{(8-1)}=\frac{1}{7}\]
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