Integral Calculus Question 304
Question: $ \int_{{}}^{{}}{{{(\tan x-\cot x)}^{2}}\ dx=} $
Options:
A) $ \tan x+\cot x+c $
B) $ \sec x\tan x+c $
C) $ \cos ecx\cot x+c $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{{{(\tan x-\cot x)}^{2}}dx}=\int_{{}}^{{}}{({{\tan }^{2}}x+{{\cot }^{2}}x-2),dx} $ $ =\int_{{}}^{{}}{{{\sec }^{2}}x,dx}+\int_{{}}^{{}}{cose{c^{2}}x,dx}-\int_{{}}^{{}}{4,dx} $ $ =\tan x-\cot x-4x+c. $
BETA
