A) \[\frac{1}{2}\]
B) 1
C) \[\infty \]
D) zero
E) none of these
Correct Answer: C
Solution :
\[\because \]\[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}dx\] \[\therefore \]\[{{I}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x(1+{{\tan }^{2}}x)dx}\] Now, \[{{I}_{n}}+{{I}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x(1+{{\tan }^{2}}x)dx}\] \[=\int_{0}^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n}}x\,dx}\] Let \[tan\text{ }x=t\] \[\Rightarrow \]\[{{\sec }^{2}}xdx=dt\] \[\therefore \]\[{{I}_{n}}+{{I}_{n+2}}=\int_{0}^{1}{{{t}^{n}}dt}\] \[=\left[ \frac{{{t}^{n+1}}}{n+1} \right]_{0}^{1}=\frac{1}{n+1}\] Hence, \[\underset{n\to \infty }{\mathop{\lim }}\,[{{I}_{n}}+{{I}_{n+2}}]=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{1+\frac{1}{n}}=1\]
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