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