Integral Calculus Question 244
Question: If $ I_{n}=\int\limits_0^{\frac{\pi }{4}}{{{\tan }^{n}}x,dx} $ then what is $ I_{n}+{I_{n-2}} $ equal to?
Options:
A) $ \frac{1}{n} $
B) $ \frac{1}{(n-1)} $
C) $ \frac{n}{(n-1)} $
D) $ \frac{1}{(n-2)} $
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Answer:
Correct Answer: B
Solution:
[b] Let $ I_{n}=\int\limits_0^{\pi /4}{{{\tan }^{n}}xdx} $ Consider, $ I_{n}+{I_{n-2}}=\int\limits_0^{\pi /4}{{{\tan }^{n}}xdx+\int\limits_0^{\pi /4}{{{\tan }^{n-2}}xdx}} $ $ =\int\limits_0^{\pi /4}{{{\tan }^{n-2}}x({{\tan }^{2}}x+1)dx} $ $ =\int\limits_0^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n-2}}xdx} $ Put $ \tan x=t $ $ {{\sec }^{2}}xdx=dt $ when $ x=0 $ then $ t=0 $ and when $ x=\frac{\pi }{4},t=1 $
$ \therefore I_{n}+{I_{n-2}}=\int\limits_0^{1}{{t^{n-2}}dt} $ $ =. \frac{{t^{n-2+1}}}{n-2+1} |_0^{1}. =\frac{{t^{n-1}}}{n-1} |_0^{1}=\frac{1}{n-1}[1-0]=\frac{1}{n-1} $
BETA
