A) 1/4
B) 1/2
C) 1/8
D) 4
E) 6
Correct Answer: D
Solution :
\[\because \]\[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}dx\] \[\therefore \] \[{{I}_{3}}+{{I}_{5}}=\int_{0}^{\pi /4}{({{\tan }^{3}}x+{{\tan }^{5}}x)}dx\] \[=\int_{0}^{\pi /4}{{{\tan }^{3}}x{{\sec }^{2}}xdx}\] \[=\left[ \frac{{{\tan }^{4}}x}{4} \right]_{0}^{\pi /4}=\frac{{{\tan }^{4}}\frac{\pi }{4}}{4}=\frac{1}{4}\] Hence, \[\frac{1}{{{I}_{3}}+{{I}_{5}}}=4\]
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