A) \[\frac{1}{4}\]
B) \[\frac{1}{5}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{7}\]
Correct Answer: D
Solution :
\[{{I}_{n}}=\int_{0}^{\pi /4}{({{\sec }^{2}}\theta }-1){{\tan }^{n-2}}\theta \,d\theta \] \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\sec }^{2}}\theta {{\tan }^{n-2}}\theta \,d\theta }-\int_{0}^{\pi /2}{{{\tan }^{n-2}}\theta }\,d\theta \] \[{{I}_{n}}=\left[ \frac{{{\tan }^{n-1}}\theta }{n-1} \right]_{0}^{\pi /4}-{{I}_{n-2}}\Rightarrow {{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-1}\] Hence \[{{I}_{8}}+{{I}_{6}}=\frac{1}{8-1}=\frac{1}{7}\].
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