A) \[\frac{2}{3}\]
B) \[\frac{1}{6}\]
C) 2
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi /6}{\frac{\sin x}{{{\cos }^{3}}x}dx=\int_{0}^{\pi /6}{\,\tan x{{\sec }^{2}}x\,dx}}\] Put \[t=\tan x\Rightarrow dt={{\sec }^{2}}x\,dx,\] then we have \[I=\int_{0}^{\frac{1}{\sqrt{3}}}{t\,dt=}\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{\frac{1}{\sqrt{3}}}=\frac{1}{6}\].
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