A) \[\frac{1}{n-1}\]
B) \[\frac{1}{n+1}\]
C) \[\frac{1}{2n-1}\]
D) \[\frac{1}{2n+1}\]
Correct Answer: A
Solution :
\[{{u}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx}\]\[=\int_{0}^{\pi /4}{({{\sec }^{2}}x-1){{\tan }^{n-2}}x\,\,dx}\] \[=\int_{0}^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n-2}}x\,\,dx}-\int_{0}^{\pi /4}{{{\tan }^{n-2}}x\,\,dx}\] \[=\left[ \frac{{{\tan }^{n-1}}x}{n-1} \right]_{0}^{\pi /4}-{{u}_{n-2}}\] \[\Rightarrow {{u}_{n}}+{{u}_{n-2}}=\frac{1}{n-1}\].
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