A) \[\frac{1}{2}\]
B) 1
C) \[\infty \]
D) zero
Correct Answer: B
Solution :
\[{{I}_{n}}+{{I}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}(1+{{\tan }^{2}}x)dx\] \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x{{\sec }^{2}}x}dx\] \[=\int_{0}^{1}{{{t}^{n}}\,dt}\]where \[t=\tan x\] \[{{I}_{n}}+{{I}_{n+2}}=\frac{1}{n+1}\] \[\Rightarrow \] \[\underset{n\to \infty }{\mathop{\lim }}\,n\left[ {{I}_{n}}+{{I}_{n+2}} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,n.\frac{1}{n+1}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n\left( 1+\frac{1}{n} \right)}=1\]
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