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 $

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Answer:

Correct Answer: B

Solution:

$ \int_{{}}^{{}}{{{\tan }^{2}}x,dx}=\int_{{}}^{{}}{({{\sec }^{2}}x-1),dx}=\tan x-x+c $ .