Integral Calculus Question 400
Question: $ \int_{{}}^{{}}{\frac{1}{{{\cos }^{2}}x{{(1-\tan x)}^{2}}}dx=} $
Options:
A) $ \frac{1}{\tan x-1}+c $
B) $ \frac{1}{1-\tan x}+c $
C) $ -\frac{1}{3}\frac{1}{{{(1-\tan x)}^{3}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{1}{{{\cos }^{2}}x{{(1-\tan x)}^{2}}}dx=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{{{(\tan x-1)}^{2}}}}} $ Put $ \tan x-1=t\Rightarrow {{\sec }^{2}}x,dx=dt, $ then it reduces to $ \int_{{}}^{{}}{\frac{1}{t^{2}},dt}=\frac{-1}{\tan x-1}+c=\frac{1}{1-\tan x}+c. $
BETA
