Integral Calculus Question 256
Question: $ \int_{{}}^{{}}{\frac{\cot x\tan x}{{{\sec }^{2}}x-1}}\ dx= $
Options:
A) $ \cot x-x+c $
B) $ -\cot x+x+c $
C) $ \cot x+x+c $
D) $ -\cot x-x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{\cot x\tan x}{{{\sec }^{2}}x-1},dx=\int_{{}}^{{}}{\frac{1}{{{\tan }^{2}}x},dx=\int_{{}}^{{}}{{{\cot }^{2}}x,dx}}} $ $ =\int_{{}}^{{}}{(cose{c^{2}}x-1),dx}=-\cot x-x+c. $
BETA
