Integral Calculus Question 397
Question: $ \int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{1+\tan x}\ dx=} $
[MP PET 1987]
Options:
A) $ \log (\cos x+\sin x)+c $
B) $ \log ({{\sec }^{2}}x)+c $
C) $ \log (1+\tan x)+c $
D) $ -\frac{1}{{{(1+\tan x)}^{2}}}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ t=1+\tan x\Rightarrow dt={{\sec }^{2}}x,dx $ $ \int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{1+\tan x}},dx=\int_{{}}^{{}}{\frac{1}{t},dt=\log t+c} $ $ =\log (1+\tan x)+c. $
BETA
