Definite Integration Question 68
Question: If $ I _{n}=\int_0^{\pi /4}{{{\tan }^{n}}\theta d\theta ,} $ then $ I_8+I_6 $ equals
[Kurukshetra CEE 1996]
Options:
A) $ \frac{1}{4} $
B) $ \frac{1}{5} $
C) $ \frac{1}{6} $
D) $ \frac{1}{7} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I _{n}=\int_0^{\pi /4}{({{\sec }^{2}}\theta }-1){{\tan }^{n-2}}\theta d\theta $
$ I _{n}=\int_0^{\pi /4}{{{\sec }^{2}}\theta {{\tan }^{n-2}}\theta d\theta }-\int_0^{\pi /2}{{{\tan }^{n-2}}\theta }d\theta $
$ I _{n}=[ \frac{{{\tan }^{n-1}}\theta }{n-1} ]_0^{\pi /4}-{I _{n-2}}\Rightarrow I _{n}+{I _{n-2}}=\frac{1}{n-1} $
Hence $ I_8+I_6=\frac{1}{8-1}=\frac{1}{7} $ .
BETA
