Integral Calculus Question 181
Question: If $ u_{n}=\int_0^{\pi /4}{{{\tan }^{n}}\theta }d\theta $ then $ u_{n}+{u_{n-2}} $ is:
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:
[a] Given: $ u_{n}=\int\limits_0^{\pi /4}{{{\tan }^{n}}\theta d\theta } $ $ =\int\limits_0^{\pi /4}{{{\tan }^{2}}\theta {{\tan }^{n-2}}\theta d\theta } $ $ =\int\limits_0^{\pi /4}{(sec^{2}\theta -1)ta{n^{n-2}}\theta d\theta } $ $ =\int\limits_0^{\pi /4}{sec^{2}\theta {{\tan }^{n-2}}\theta d\theta -\int\limits_0^{\pi /4}{{{\tan }^{n-2}}\theta d\theta }} $ $ =\int\limits_0^{\pi /4}{sec^{2}\theta {{\tan }^{n-2}}\theta d\theta -{u_{n-2}}} $
$ \Rightarrow u_{n}+{u_{n-2}}=\int\limits_0^{\pi /4}{{{\sec }^{2}}\theta {{\tan }^{n-2}}\theta d\theta } $ $ =. \frac{{{\tan }^{n-1}}\theta }{n-1} |_0^{\pi /4}=\frac{1}{n-1} $
BETA
