Integral Calculus Question 315
Question: $ \int_{{}}^{{}}{\frac{1-\tan x}{1+\tan x}\ dx=} $
[MP PET 1994]
Options:
A) $ \log \sec ( \frac{\pi }{4}-x )+c $
B) $ \log \cos ( \frac{\pi }{4}+x )+c $
C) $ \log \sin ( \frac{\pi }{4}+x )+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{1-\tan x}{1+\tan x},dx}=\int_{{}}^{{}}{\tan ( \frac{\pi }{4}-x )},dx $ $ =\int_{{}}^{{}}{\frac{\sin ( \frac{\pi }{4}-x )}{\cos ( \frac{\pi }{4}-x )}},dx=\log \cos ( \frac{\pi }{4}-x )+c $ $ =\log \sin ( \frac{\pi }{4}+x )+c $ .
BETA
