Definite Integration Question 110
Question: If $ u _{n}=\int_0^{\pi /4}{{{\tan }^{n}}xdx,} $ then $ u _{n}+{u _{n-2}}= $
[UPSEAT 2002]
Options:
A) $ \frac{1}{n-1} $
B) $ \frac{1}{n+1} $
C) $ \frac{1}{2n-1} $
D) $ \frac{1}{2n+1} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ u _{n}=\int_0^{\pi /4}{{{\tan }^{n}}xdx} $
$ =\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}-{u _{n-2}} $
$ \Rightarrow u _{n}+{u _{n-2}}=\frac{1}{n-1} $ .
BETA
