A) 1
B) \[n-1\]
C) \[\frac{1}{n-1}\]
D) None of these
Correct Answer: A
Solution :
We have \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta \,d\theta }\] Now, \[{{I}_{n+1}}=\int_{0}^{\pi /4}{{{\tan }^{n+1}}}\theta \,d\theta \] \[=\int_{0}^{\pi /4}{{{\tan }^{n-1}}\theta {{\tan }^{2}}\theta \,d\theta }\] \[=\int_{0}^{\pi /4}{{{\tan }^{n-1}}\theta }(se{{c}^{2}}\theta -1)d\theta \] \[\Rightarrow \] \[{{I}_{n+1}}=\int_{0}^{\pi /4}{{{\tan }^{n-1}}\theta {{\sec }^{2}}\theta \,d\theta }\] \[-\int_{0}^{\pi /4}{{{\tan }^{n-1}}\theta }\,d\theta \] \[\Rightarrow \] \[{{I}_{n+1}}=\int_{0}^{\pi /4}{{{\tan }^{n-1}}}\theta {{\sec }^{2}}\theta d\theta -{{I}_{n-1}}\] ?(i) Now, \[\int_{0}^{\pi /4}{{{\tan }^{n-1}}\theta }{{\sec }^{2}}\theta \,d\theta \] Putting \[\tan \theta ={{t}_{1}}\Rightarrow {{\sec }^{2}}\theta \,d\theta =dt\] \[=\int_{0}^{1}{{{t}^{n-1}}dt}\] \[=\left[ \frac{{{t}^{n}}}{n} \right]_{0}^{1}=\left( \frac{1}{n}-0 \right)\] \[=\frac{1}{n}\] Putting this value in Eq. (i), we get \[{{I}_{n+1}}=\frac{1}{n}-{{I}_{n-1}}\] \[\Rightarrow \] \[{{I}_{n+1}}+{{I}_{n-1}}=\frac{1}{n}\] \[\Rightarrow \] \[n({{I}_{n-1}}+{{I}_{n+1}})=1\]
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