Integral Calculus Question 362
Question: $ \int_{{}}^{{}}{{{\tan }^{2}}x\ dx} $ is equal to
[SCRA 1996]
Options:
A) $ \tan x+x+c $
B) $ \tan x-x+c $
C) $ \sec x+x+c $
D) $ \sec x-x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{\tan }^{2}}x,dx}=\int_{{}}^{{}}{({{\sec }^{2}}x-1),dx}=\tan x-x+c $ .