A) \[\frac{1}{9}\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{7}\]
D) 9
Correct Answer: A
Solution :
Given integral is, \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx}\] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n-2}}x.{{\tan }^{2}}x\,\,dx}\] \[=\int_{0}^{\pi /4}{{{\tan }^{n-2}}x.\,(se{{c}^{2}}x-1)dx}\] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n-2}}x.\,se{{c}^{2}}x\,\,dx}\] \[-\int_{0}^{\pi /4}{{{\tan }^{n-2}}x\,\,dx}\] \[{{I}_{n}}=\int_{0}^{1}{{{t}^{n-2}}dt-\int_{0}^{\pi /4}{{{\tan }^{n-2}}x\,dx}}\] \[{{I}_{n}}=\left[ \frac{{{t}^{n-1}}}{n-1} \right]_{0}^{1}-{{I}_{n-2}}\] \[\left\{ \begin{align} & \text{Put tan x =t } \\ & \Rightarrow \,\,\text{se}{{\text{c}}^{2}}\,x\,dx=dt \\ \end{align} \right\}\] \[{{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-1}\] Put \[n=10,\]we get \[{{I}_{10}}+{{I}_{8}}=\frac{1}{9}\]
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