Definite Integration Question 449
Question: $ I _{n}=\int _{0}^{\pi /4}{{{\tan }^{n}}xdx} $ , then $ \underset{n-\infty }{\mathop{\lim }}n[I _{n}+{I _{n-2}}] $ equals
[AIEEE 2002]
Options:
A) 1/2
B) 1
C) $ \infty $
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
$ I _{n}=\int_0^{\pi /4}{{{\tan }^{n}}x}dx=\int_0^{\pi /4}{({{\sec }^{2}}x-1){{\tan }^{n-2}}xdx} $
$ =\int_0^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n-2}}}xdx-\int_0^{\pi /4}{{{\tan }^{n-2}}}xdx $
$ =[ \frac{{{\tan }^{n-1}}x}{n-1} ]_0^{\pi /4}-{I _{n-2}} $
therefore $ I _{n}+{I _{n-2}}=\frac{1}{n-1} $
Now, $ \underset{n\to \infty }{\mathop{\lim }}n[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 $ .
BETA
