Integral Calculus Question 77
Question: $ \int_{{}}^{{}}{\frac{\sec x\ dx}{\sqrt{\cos 2x}}}= $
Options:
A) $ {{\sin }^{-1}}(\tan x) $
B) $ \tan x $
C) $ {{\cos }^{-1}}(\tan x) $
D) $ \frac{\sin x}{\sqrt{\cos x}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{\sec x,dx}{\sqrt{\cos 2x}}}=\int_{{}}^{{}}{\frac{\sec x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}},dx $ $ =\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{\sqrt{1-{{\tan }^{2}}x}}} $ {Multiplying $ N’r $ and $ D’r $ by $ \sec x} $ Now putting $ \tan x=t\Rightarrow {{\sec }^{2}}x,dx=dt, $ we get the integral $ ={{\sin }^{-1}}t={{\sin }^{-1}}(\tan x). $ Trick : Since $ \frac{d}{dx}{{{\sin }^{-1}}(\tan x)}=\frac{{{\sec }^{2}}x}{\sqrt{1-{{\tan }^{2}}x}} $ $ =\frac{{{\sec }^{2}}x.\cos x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}=\frac{\sec x}{\sqrt{\cos 2x}}. $
BETA
