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\[t=\tan x\] \[\Rightarrow \] \[{{I}_{n}}+{{I}_{n+2}}=\frac{1}{n+1}\] \[\therefore \] \[\underset{n\to \infty }{\mathop{\lim }}\,n[{{I}_{n}}+{{I}_{n+2}}]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,n\cdot \frac{1}{n+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n\left( 1+\frac{1}{n} \right)}=1\]You need to login to perform this action.
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