Definite Integration Question 381
Question: $ \int_0^{\pi /6}{\frac{\sin x}{{{\cos }^{3}}x}dx=} $
[SCRA 1979]
Options:
A) $ \frac{2}{3} $
B) $ \frac{1}{6} $
C) 2
D) $ \frac{1}{3} $
Show Answer
Answer:
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}}xdx}} $
Put $ t=\tan x\Rightarrow dt={{\sec }^{2}}xdx, $ then we have $ I=\int_0^{\frac{1}{\sqrt{3}}}{tdt=}[ \frac{t^{2}}{2} ]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} $ .
BETA
